NCERT Solutions for Class 12 Maths Chapter 10 - Vectors - Miscellaneous Exercise

Access Exercises of Class 12 Maths Chapter 10 – Vectors

Miscellaneous Exercise

1. Write down a unit vector in XY-plane, making an angle of 30° in anticlock-wise direction with the positive direction of X-axis.

Sol. Let OP make 30° with X-axis, 60° with Y-axis and 90° with Z-axis and it lies in XY-plane.

$$\therefore\space\text{Direction cosines of}\space\vec{\text{OP}}\\\text{are cos 30°, cos 60° and}\\\text{cos 90\degree i.e.,}\frac{\sqrt{3}}{2},\frac{1}{2},0.\space\text{Thus,}\space\vec{\text{OP}}\space\text{is}\\\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\\|\vec{\text{OP}}| = \sqrt{\bigg(\frac{\sqrt{3}}{2}\bigg)^{2} + \bigg(\frac{1}{2}\bigg)}\\=\sqrt{\frac{3}{4}+\frac{1}{4}}\\=\sqrt{1} = 1$$

which is required unit vector of XY-plane.

2. Find the scalar components and magnitude of the vector joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂).

Sol. The vector joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) can be obtained by,

$$\vec{\text{OP}} = \text{P.V. of Q – P.V. of P}\\ =(x_2\hat{i} + y_2\hat{j} + z_2\hat{k}) - (x_1\hat{i} +y_1\hat{j}+z_1\hat{k})\\=(x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}\\\therefore\space\text{Scalar components of}\space\vec{\text{PQ}}\\\space\text{are}\space x_2-x_1, y_2-y_1,z_2-z_1.\\\therefore\space\text{Magnitude of}|\vec{\text{PQ}}|\\=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}+(z_2-z_1)^{2}}$$

3. A girls walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl's displacement from her initial point of departure.

Sol. Let O and B be the initial and final positions of the girl respectively.

Then, the girl's position can be shown as in the figure.

$$\text{Now, we have}\space \vec{\text{OA}} = 4\hat{i}\\\vec{\text{AB}} =\hat{i}|\vec{\text{AB}}|\text{cos} 60\degree + \hat{j}|\vec{\text{AB}}|\text{sin}60\degree\\(\vec{\text{AB}}\space \text{cos} \space60\degree\text{is component}\space\vec{\text{AB}},\\\text{along X-axis and}\space \vec{\text{AB}\space \text{sin 60\degree is component}}\\\text{of\space}\vec{\text{AB} = \text{along Y-axis})}\\ = \hat{i}\space 3×\frac{1}{2}+\hat{j}\space 3×\frac{\sqrt{3}}{2}\\=\frac{3}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}$$

By the triangle law of vector addition, we have

$$\vec{\text{OB}} =\vec{\text{OA}} + \vec{\text{AB}}\\=(-4\hat{i}) + \bigg(\frac{3}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}\bigg)\\= \bigg(-4 + \frac{3}{2}\bigg)\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}\\ =\bigg(\frac{-8+3}{2}\bigg)\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}\\=\frac{-5}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j} $$

Hence, the girl's displacement from her initial point of departure is

$$\frac{-5}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}. $$

$$\textbf{4.\space If}\space \vec{\textbf{a}} \space\textbf{=}\space\vec{\textbf{b}} +\vec{\textbf{c}}\textbf{,}\space\textbf{then is it true that}\\\textbf{|}\vec{\textbf{a}}\textbf{|} \textbf{=}\textbf{|}\vec{\textbf{b}}\textbf{|} \textbf{+}|\vec{\textbf{c}}|\space \textbf{?}\space\textbf{Justify you answer.}\\\textbf{Sol.\space}\text{In triangle ABC, let}\space\\\vec{\text{CB}} =\vec{a},\vec{\text{CA}} =\vec{b}\space\text{and}\space \vec{\text{AB}} = \vec{c}$$

(as shown in the following figure)

Now , by the triangle law of addition, we have $$\vec{a} = \vec{b}+\vec{c}.$$

It is clearly known that

$$|\vec{a}|,|\vec{b}|\space\text{and}|\vec{c}|\space\\\text{represent the side of}\Delta \text{ABC.}$$

5. Find the value of x for which

$$\textbf{x}(\hat{\textbf{i}}+\hat{\textbf{j}} + \hat{\textbf{k}})\space\textbf{is a unit vector.}$$

$$\textbf{Sol.\space}\text{Given,\space}x(\hat{i} +\hat{j} + \hat{k})\\\space\text{is a unit vector, therefore,}\\|x(\hat{i} + \hat{j}+\hat{k})|=1\\\Rarr\space |x\hat{i} +x\hat{j} + x\hat{k}|=1\\\Rarr\space|\sqrt{x^{2}+x^{2}+x^{2}}|=1\\\Rarr\space \sqrt{3x^{2}}=1\\\Rarr\space \pm\sqrt{3x}=1\\\Rarr\space x = \pm\frac{1}{\sqrt{3}}\\\text{Hence, the required values of x are}\\\pm\frac{1}{\sqrt{3}}.$$

6. Find a vector of magnitude 5 units and parallel to the resultant of the vectors

$$\vec{\textbf{a}} \textbf{= 2}\hat{\textbf{i}} \textbf{+ 3}\hat{\textbf{j}}\textbf{-}\hat{\textbf{k}}\space\textbf{and}\space \vec{\textbf{b}} \textbf{=}\hat{\textbf{i}}\textbf{-2}\hat{\textbf{j}}\textbf{+}\hat{\textbf{k}}\\\textbf{Sol.}\space\text{Given vectors }\vec{a} =2\hat{i} +3\hat{i}-\hat{k}\\\text{and}\space\vec{b} = \hat{i} - 2\hat{j} +\hat{k}\\\text{Let}\space\vec{c}\space\text{be the resultant of}\space\vec{a}\space\text{and}\space\vec{b}.\\\therefore\space\vec{c} = \vec{a} +\vec{b}\\=(2\hat{i}+3\hat{j}-\hat{k}) + (\hat{i}-2\hat{j}+\hat{k})\\\Rarr\space \vec{c} = 3\hat{i}+\hat{j}+0\hat{k}\\\text{Comparing with}\space\vec{\text{X}} = x\hat{i}+y\hat{j}+z\hat{k}\\\therefore\space |\vec{c}| = \sqrt{x^{2}+y^{2}+z^{2}}\\=\sqrt{3^{2}+1^{2}}\\=\sqrt{9+1}=\sqrt{10}\\\therefore\space \text{Unit vector in the direction of}\\\hat{c} = \frac{\vec{c}}{|\vec{c}|} =\frac{3\hat{i} +\hat{j}}{\sqrt{10}}$$

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors

$$\vec{a}\space\text{and}\space\vec{b}\space\text{is}\\\pm 5\hat{c} = \pm 5\frac{1}{\sqrt{10}}(3\hat{i} +\hat{j}) \\=\pm\frac{3\sqrt{10}}{2}\hat{i}\pm\frac{\sqrt{10}}{2}\hat{j}$$

$$\textbf{7.\space If}\space\vec{\textbf{a}}\space\textbf{=}\space\hat{\textbf{i}}\textbf{+}\hat{\textbf{j}}\textbf{+} \hat{\textbf{k}},\space \vec{\textbf{b}} \textbf{= 2}\hat{\textbf{i}} \textbf{-}\textbf{j} \textbf{+ 3}\hat{\textbf{k}}\space\\\textbf{and}\space\vec{\textbf{c}} \space\textbf{=} \space\hat{\textbf{i}}\textbf{-2}\hat{\textbf{j}}\textbf{+}\hat{\textbf{k}}\space\textbf{find a unit}\\\textbf{vector parallel to the vector}\space\\ \textbf{2}\vec{\textbf{a}}-\vec{\textbf{b}} \textbf{+ 3}\vec{\textbf{c}}.$$

Sol. We have,

$$\vec{a} =\hat{i} + \hat{j} +\hat{k},\\\vec{b} = 2\hat{i} -\hat{j} + 3\hat{k}\space\text{and}\space\vec{c} =\hat{i} -2\hat{j}+\hat{k}\\\text{Let}\space \vec{v} = 2\vec{a} - \vec{b} + 3\vec{c}\\=2(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+3\hat{k})\\ +\space 3(\hat{i}-2\hat{j}+\hat{k})\\=3\hat{i} -3\hat{j}+2\hat{k}\\\text{Now,}\space |\vec{v}| = 3\hat{i} + 3\hat{j}+ 2\hat{k}\\=\sqrt{3^{2}+(-3)^{2}+2^{2}}\\=\sqrt{22}\\\text{Hence, the unit}\space\text{vector angle}\\\vec{v}\space\text{is}\\\hat{v} =\frac{\vec{v}}{|\vec{v}|} =\frac{3\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{22}}$$

$$=\bigg(\frac{3}{\sqrt{22}}\hat{i} -\frac{3}{\sqrt{22}}\hat{j} + \frac{2}{\sqrt{22}}\hat{k}\bigg)$$

8. Show that the points A(1, – 2, – 8), B(5, 0, – 2) and C(11, 3, 7) are collinear and find the ratio in which B divides AC.

Sol. The given points are A(1, – 2, – 8), B(5, 0, – 2) and C(11, 3, 7).

$$\vec{\text{AB}} =\text{P.V. of B – P.V. of A}\\ =(5\hat{i} + 0\hat{j}-2\hat{k})-(\hat{i}-2\hat{j}-8\hat{k})\\=(5-1)\hat{i}+(0+2)\hat{j}+(-2+8)\hat{k}\\= 4\hat{i}+2\hat{j}+6\hat{k}\\|\vec{\text{AB}}| = \sqrt{4^{2}+2^{2}+6^{2}}\\=\sqrt{16+4+36}\\=\sqrt{56} = 2\sqrt{14}\\\vec{\text{BC}} = \text{P.V. of C – P.V. of B}\\=(11\hat{i} +3\hat{j}+7\hat{k}) - (5\hat{i} + 0\hat{j}-2\hat{k})\\=(11-5)\hat{i}+(3-0)\hat{j}+(7+2)\hat{k}\\= 6\hat{i}+3\hat{j}+9\hat{k}\\|\vec{\text{BC}}| =\sqrt{6^{2} + 3^{2}+9^{2}}\\=\sqrt{36 + 9 + 81}\\=\sqrt{126}$$

$$= 3\sqrt{14}\space\\\vec{\text{AC}} = \text{P.V. of C – P.V. of A}\\=(11\hat{i} + 3\hat{j}+7\hat{k}) -(\hat{i}-2\hat{j}-8\hat{k})\\=(11-1)\hat{i} + (3+2)\hat{j}+(7+8)\hat{k}\\=10\hat{i}+5\hat{j}+15\hat{k}\\|\vec{\text{AC}}| = \sqrt{10^{2}+5^{2}+15^{2}}\\ =\sqrt{100+25 + 225}\\=\sqrt{350} =5\sqrt{14}\\\therefore\space |\vec{\text{AC}}| =|\vec{\text{AB}}| +|\vec{\text{BC}}|$$

Thus, the given points A, B and C are collinear.

Let P be the point (on the line AC) which divides [AC] in the ratio l : 1, then P.V. of the point

$$\text{P} =\frac{\lambda×\text{P.V. of C}+1×\text{P.V. of A}}{\lambda +1}$$

$$=\frac{1}{\lambda+1}\lbrack(11\hat{i}+3\hat{j}+7\hat{k})+\\1(\hat{i}-2\hat{j}-8\hat{k})\rbrack\\=\bigg(\frac{11\lambda+1}{\lambda+1}\bigg)\hat{i}+\bigg(\frac{3\lambda-2}{\lambda-1}\bigg)\hat{j}+\\\bigg(\frac{7\lambda-8}{\lambda+1}\bigg)\hat{k}\\= 5\hat{i}+0\hat{j}-2\hat{k}\\\Rarr\space \frac{11\lambda+1}{\lambda+1}=5,\frac{3\lambda-2}{\lambda+1}=0\\\text{and}\space\frac{7\lambda-8}{\lambda+1}=-2$$

$$\Rarr\space 11\lambda+1 =5\lambda+5,\space 3\lambda=2,\\ 7\lambda-8 =-2\lambda-2\\\Rarr\space 6\lambda=4\lambda\\\Rarr\space \frac{2}{3}, 9\lambda =6\\\Rarr\space \lambda =\frac{2}{3}$$

Hence, A, B, C are collinear and B divides [AC] in

$$\text{the ratio}\space\frac{2}{3}\space :\text{i.e., 2:3.}$$

9. Find the position vector of a point R which divides the line joining two points P and Q whose position vector are

$$\textbf{(2}\vec{\textbf{a}} \textbf{+} \vec{\textbf{b}}\textbf{)}\space\textbf{and}\space \textbf{(}\vec{\textbf{a}} - 3\vec{\textbf{b}}\textbf{)}$$

externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.

Sol. It is given that

$$\vec{\text{OP}} =\space 2\vec{a} + \vec{b},\vec{\text{OQ}} = \vec{a} - 3 \vec{b}$$

If a point divides the line joining point P and Q externally in the ratio m : n, then position vector of the point is

$$\frac{m(\vec{\text{PV}} \text{of Q}) - n(\vec{\text{P.V.}} \text{of P})}{\text{m-n}}$$

It is given that point R divides a line segment joining two points P and Q externally in the ratio 1 : 2. Then, on using the section formula, we get position vector of point

$$\text{R} = \frac{(\vec{a} - 3\vec{b})×1 - (2\vec{a} + \vec{b})×2}{1-2}\\=\frac{\vec{a} - 3\vec{b}-4\vec{a}-2\vec{b}}{\normalsize-1}\\=\frac{-3\vec{a} - 5\vec{b}}{-1} = 3\vec{a}+5\vec{b}\\\text{Now , position vector of mid-point of}\space\\\vec{\text{PQ}}\\=\frac{\vec{\text{OQ}}+\vec{\text{OR}}}{2}\\=\frac{(3\vec{a}+5\vec{b}) + (\vec{a} - 3\vec{b})}{2}\\=\frac{4\vec{a}+2\vec{b}}{2}\\= 2\vec{a} + \vec{b}$$

Also, the position vector of point

$$\text{P = 2}\vec{\text{a}} + \vec{\text{b}}, $$

which shows that P is mid-point of line segment RQ.

10. The two adjacent sides of a parallelogram are

$$\textbf{2}\hat{\textbf{i}}\textbf{- 4}\hat{\textbf{j}}\textbf{+ 5}\hat{\textbf{k}}\space\textbf{and}\space \vec{\textbf{b}} \textbf{=}\hat{\textbf{i}}\textbf{-2}\hat{\textbf{j}}\textbf{- 3}\hat{\textbf{k}}\textbf{.}$$Find the unit vector parallel to its diagonal. Also, find its area.

Sol. Adjacent sides of a parallelogram are given as

$$\vec{a} = 2\hat{i}-4\hat{j}+5\hat{k}\space\text{and}\space\vec{b} =\hat{i} - 2\hat{j}-3\hat{k}. $$

Then, the diagonal of a parallelgoram is given by

$$\vec{v} = \vec{a}+\vec{b}.$$

(∵ From the figure, it is clear that resultant of adjacent sides of a parallelogram is given by the diagonal)

$$\therefore\space \vec{v} = 2\hat{i}-4\hat{j}+5\hat{k}+\hat{i}-2\hat{j}-3\hat{k}\\=(2+1)\hat{i}+(-4-2)\hat{j}+(5-3)\hat{k}\\=3\hat{i}-6\hat{j}+2\hat{k}\\\text{Comparing with}\space \\\vec{\text{X}} = x\hat{i}+y\hat{j}+z\hat{k},\space\text{we get}$$

x = 3, y = – 6, z = 2

$$\therefore\space |\vec{\text{V}}| = \sqrt{x^{2}+y^{2}+z^{2}}\\=\sqrt{(3)^{2}+(-6)^{2}+(2)^{3}}\\=\sqrt{9+36+4} =\sqrt{49} =7$$

Thus, the unit vector parallel to the diagonal is

$$\frac{\vec{v}}{|\vec{v}|} =\frac{3\hat{i} -6\hat{j}+2\hat{k}}{7}\\=\frac{3}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}$$

Also, area of parallelogram ABCD,

$$|\vec{a}×\vec{b}| = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 2 &-4 &5\\ 1&-2 &-3\end{vmatrix}\\=|\hat{i}(12 + 10) -\hat{j}(-6-5) + \hat{k}(-4+4)|\\=|22\hat{i} + 11\hat{j} + 10\hat{k}|\\=\sqrt{(22)^{2}+(11)^{2}+0^{2}}\\=\sqrt{(11)^{2}(2^{2}+1^{2})}\\= 11\sqrt{5}\space\text{sq unit}$$

11. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are

$$\pm\bigg(\frac{\textbf{1}}{\sqrt{\textbf{3}}}\textbf{,}\frac{\textbf{1}}{\sqrt{\textbf{3}}}\textbf{,}\frac{\textbf{1}}{\sqrt{\textbf{3}}}\bigg)\textbf{.}$$

Sol. Let a vector be equally inclinded to OX, OY and OZ and let if make an angle a with each of these three, then the direction cosines are cos a, cos a and cos α.

$$\Rarr\space \text{cos}^{2}\alpha + \text{cos}^{2}\alpha + \text{cos}^{2}\alpha=1\\(\because\space l^{2}+m^{2}+n^{2}=1)\\\Rarr\space \text{3 cos}^{2}\alpha=1\\\Rarr\space \text{cos}^{2}\alpha = \frac{1}{3}\\\Rarr\space \text{cos}\space\alpha = \pm\frac{1}{\sqrt{3}}$$

Hence, the directon cosines of the vector which is equally inclined to the axes are either

$$\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\space\text{or}\space-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}.$$

$$\textbf{12.\space}\textbf{Let}\space \vec{\textbf{a}} \space\textbf{=}\space \hat{\textbf{i}} \textbf{+ 4}\hat{\textbf{j}} \textbf{+ 2}\hat{\textbf{k}},\space \vec{\textbf{b}} \textbf{= 3}\hat{\textbf{i}}\textbf{-2}\textbf{j} \textbf{+ 7}\hat{\textbf{k}}\\\textbf{and}\space\vec{\textbf{c}} \textbf{= 2}\hat{\textbf{i}}\textbf{-}\hat{\textbf{j}} \textbf{+ 4}\hat{\textbf{k}}.\space\textbf{Find a vector}\\\vec{a}\space \textbf{which is perpendicular to both}\\\vec{\textbf{a}}\space\textbf{and}\space\vec{\textbf{b}}\space\textbf{and}\space\vec{\textbf{c}}.\vec{\textbf{d}}\textbf{ = 15}.$$

Sol. The vector which is perpendicular to both

$$\vec{a}×\vec{b}\space\text{must be parallel to}\space\vec{a}×\vec{b}.\\\text{Now,}\space \vec{a}×\vec{b}\begin{vmatrix}\hat{i} &\hat{j}&\hat{k}\\ 1 &4 &2\\3 &-2 &7\end{vmatrix}\\ = \hat{i(28+4)} -\hat{j}(7-6) + \hat{k}(-2-12)\\= 32\hat{i} -\hat{j}-14\hat{k}\\\text{Let}\space \vec{d} = \lambda(\vec{a}×\vec{b})\\=\lambda(32\hat{i} -\hat{j}-14\hat{k})\\\text{Also,}\space \vec{\text{c}}.\vec{\text{d}} = 15\\\Rarr\space (2\hat{i} -\hat{j} + 4\hat{k}).\lambda(32\hat{i} -\hat{j}-14\hat{k})=15\\\Rarr\space 2×(32\lambda) + (\normalsize-1)×(-\lambda) + 4×(-14\lambda) = 15\\\Rarr\space 64\lambda + \lambda - 56\lambda = 15\\\Rarr\space 9\lambda = 15 $$

$$\Rarr\lambda\space = \frac{15}{9}=\frac{15}{3}\\\therefore\space\text{Required vector,}\space \vec{d} = \frac{5}{3}(32\hat{i} -\hat{j}-14\hat{k})$$

Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.

$$\therefore\space |\vec{a}| \lt |\vec{b}|+|\vec{c}|\\\text{Hence, it is not true that}\\|\vec{a}| =|\vec{b}| + |\vec{c}|.$$

13. The scalar product of the vector

$$\hat{\textbf{i}} + \hat{\textbf{j}} + \hat{\textbf{k}}\space\textbf{with a unit vector along}\\\textbf{the sum of vectors}\space \textbf{2}\hat{\textbf{i}} \textbf{+ 4}\hat{\textbf{j}} \textbf{- 5}\hat{\textbf{k}}\space\textbf{and}\\\lambda\hat{\textbf{i}}\textbf{ + 2}\hat{\textbf{j}} \textbf{+ 3}\hat{\textbf{k}}\space\textbf{is equal to one.}\\\textbf{Find the value of}\space\lambda.$$

$$\textbf{Sol.\space}\text{Let}\space \vec{a} = \hat{i} +\hat{j}+ \hat{k},\\\vec{b} = 2\hat{i} + 4\hat{j}-5\hat{k}\space\text{and}\space\\\vec{c} =\lambda \hat{i} + 2\hat{j}+3\hat{k}\\\text{Now,\space}\vec{b} + \vec{c} = 2\hat{i} + 4\hat{i}-5\vec{k}+\lambda\hat{i}+2\hat{j}+3\hat{k}\\ = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}\\\therefore\space|\vec{b} +\vec{c}| = \\\sqrt{(2 + \lambda)^{2} + (6)^{2} + (-2)^{2}}\\=\sqrt{4 +\lambda^{2} + 4\lambda + 36+4}\\=\sqrt{\lambda^{2} + 4\lambda +44}\\\text{The unit vector along}\space (\vec{b} +\vec{c}),\text{i.e.,}\\\frac{\vec{b}+\vec{c}}{|\vec{b} + \vec{c}|} = \frac{(2 + \lambda)\hat{i} +6\hat{j} -4\hat{k}}{\sqrt{\lambda^{2} + 4\lambda +44}}$$

$$\text{Scalar product}\space (\hat{i} +\hat{j} +\hat{k})\space\text{with}\\\text{this unit vector is 1.}\\\therefore\space (\hat{i} + \hat{j} + \hat{k}).\frac{\vec{b}.\vec{c}}{|\vec{b} + \vec{c}|} = 1\\\Rarr\space (\hat{i} + \hat{j} + \hat{k}).\frac{(2 +\lambda)\hat{i} + 6\hat{j}-2\hat{k}}{\sqrt{\lambda^{2}+4\lambda +44}}=1\\\Rarr\space\frac{1(2 +\lambda) +1(6) + 1(-2)}{\sqrt{\lambda^{2} + 4\lambda +44}}=1\\\Rarr\space \frac{(2 + \lambda) + 6-2}{\sqrt{\lambda^{2} +4\lambda + 44}} = 1\\\Rarr\space\lambda + 6 =\sqrt{\lambda^{2} +4\lambda +44}\\\Rarr\space (\lambda + 6)^{2} = \lambda^{2} + 4\lambda+44\\\Rarr\space \lambda^{2} + 12\lambda+36 =\lambda^{2} + 4\lambda+44$$

$$\Rarr\space 8\lambda = 8\\\Rarr\space \lambda = 1.\\\text{Hence, the value of}\space\lambda\space\text{is 1.}$$

$$\textbf{14.\space}\textbf{If}\space\vec{\textbf{a}},\vec{\textbf{b}},\vec{\textbf{c}}\space\textbf{are mutually perpendicular}\\\textbf{of equal magnitudes, show that the}\\\textbf{vector}\space \textbf{(}\vec{\textbf{a}} + \vec{\textbf{b}} + \vec{\textbf{c}}\textbf{)}\space\textbf{is equally inclined to}\\\vec{\textbf{a}},\vec{\textbf{b}}\space\textbf{and}\space \vec{\textbf{c}.}\\\textbf{Sol.\space}\text{Given}\space\vec{a}.\vec{b} =\vec{b}.\vec{c} =\vec{c}.\vec{a} =0\\(\because\space \vec{a},\vec{b},\vec{c}\space \text{are mutually perpendicular})\\\text{It is also given that}\space|\vec{a}|= |\vec{b}|=|\vec{c}|\\(\because\space \text{All the vectors have same magnitude})\\\text{Let vector}\space (\vec{a} + \vec{b} + \vec{c})\space\text{be inclined to}\space \vec{a}, \vec{b}\\\text{and}\space\vec{c}\space\text{at angles}\space\alpha,\beta\space\text{and}\space\gamma\space\text{respectively.}$$

Then, we have

$$\text{cos}\space\alpha =\frac{(\vec{a} + \vec{b} + \vec{c}).\vec{a}}{|\vec{a} +\vec{b}+\vec{c}||\vec{a}|}\\=\frac{\vec{a}.\vec{a} + \vec{b}.\vec{a} + \vec{c}.\vec{a}}{|\vec{a} + \vec{b} +\vec{c}||\vec{a}|}\\=\frac{\vec{a}.\vec{a} +0+0}{|\vec{a} + \vec{b} + \vec{c}||\vec{a}|}\\=\frac{|\vec{a}|^{2}}{|\vec{a}+ \vec{b} + \vec{c}||\vec{a}|}\\\lbrack\because\space \vec{b}.\vec{c} =\vec{c}.\vec{a} =0\rbrack\\ = \frac{|\vec{a}|}{|\vec{a} + \vec{b} + \vec{c}|}\\\text{cos}\beta =\frac{(\vec{a}+\vec{b} +\vec{c}).\vec{b}}{|\vec{a} + \vec{b}+\vec{c}|||\vec{b}|}$$

$$ = \frac{\vec{a}.\vec{b} + \vec{b}.\vec{b} + \vec{c}.\vec{b}}{|\vec{a} + \vec{b}+ \vec{c}||\vec{b}|}\\=\frac{\vec{b}.\vec{b}+0+0}{|\vec{a} + \vec{b} +\vec{c}||\vec{b}|}\\ = \frac{|\vec{b}|^{2}}{|\vec{a} +\vec{b} + \vec{c}| |\vec{b}|}\\\lbrack\because\space \vec{a} .\vec{b} = \vec{c}.\vec{b} = 0\rbrack\\=\frac{|\vec{b}|}{|\vec{a} +\vec{b} +\vec{c}|}\\\text{cos}\space\gamma = \frac{(\vec{a}+ \vec{b} +\vec{c}).\vec{c}}{|\vec{a} +\vec{b} + \vec{c}||\vec{c}|}$$

$$= \frac{\vec{a}.\vec{c} + \vec{b}.\vec{c} +\vec{c}.\vec{c}}{|\vec{a}+\vec{b}+\vec{c}||\vec{c}|}\\=\frac{\vec{c}.\vec{c} + 0 +0}{|\vec{a} + \vec{b} +\vec{c}||\vec{c}|}\\=\frac{|\vec{c}|^{2}}{|\vec{a} +\vec{b} + \vec{c}||\vec{c}|}\\\lbrack\because\space \vec{a}.\vec{c} = \vec{b}.\vec{c} = 0\rbrack\\= \frac{|\vec{c}|}{|\vec{a} +\vec{b}+ \vec{c}|}.\\\text{Now,\space as}\space |\vec{a}| =|\vec{b}| = |\vec{c}|,\\\text{therefore, cos}\alpha = \text{cos}\beta = \text{cos}\gamma\\\therefore\space \alpha = \beta = \gamma$$

$$\text{Hence, the vector}\space(\vec{a} +\vec{b}+ \vec{c})\space\text{is}\\\text{equally inclined to}\space\vec{a},\vec{b}\space\text{and}\space \vec{c}.$$

$$\textbf{15. Prove that}\space\textbf{(}\vec{\textbf{a}} + \vec{\textbf{b}}\textbf{).}\textbf{(}\vec{\textbf{a}}+ \vec{\textbf{b}}) \textbf{=}\\\textbf{|}\vec{\textbf{a}}\textbf{|}^{\textbf{2}} \textbf{+ |}\vec{\textbf{b}}\textbf{|}^{\textbf{2}} \textbf{,}\space\textbf{if and only if}\space\vec{\textbf{a}},\vec{\textbf{b}}\space\\\textbf{are perpendicular, given}\space\vec{\textbf{a}}\neq \textbf{0},\vec{\textbf{b}}\neq \textbf{0 .}$$

Sol. Given,

$$(\vec{a} + \vec{b}).(\vec{a} + \vec{b}) = |\vec{a}|^{2} + |\vec{b}|^{2}\\\Rarr\space \vec{a}.(\vec{a} +\vec{b}) + \vec{b}.(\vec{a} +\vec{b}) = |\vec{a}|^{2} + |\vec{b}|^{2}\\\Rarr\space \vec{a}.\vec{a} + \vec{a}.\vec{b} +\vec{b}.\vec{a}+\vec{b}.\vec{b}\\=|\vec{a}|^{2}+|\vec{b}|^{2}\\\Rarr\space|\vec{a}|^{2} + 2\vec{a}.\vec{b} + |\vec{b}|^{2} =|\vec{a}|^{2} + |\vec{b}|^{2}\\2\space\vec{a}.\vec{b} = 0\\\lbrack\because\space\vec{a}.\vec{b} = \vec{b}.\vec{a}\space\text{(scalar product is commutative)}\rbrack\\\Rarr\space \vec{a}.\vec{b} = 0\\\therefore\space\vec{a}\space\text{and}\space\vec{b}\space\text{are perpendicular.}\\\lbrack(\because\space\vec{a} \neq 0, \vec{b}\neq 0)\rbrack\\\therefore\space \vec{a}\space\text{and}\space\vec{b}\space\text{are perpendicular.}$$

Choose the correct answer

$$\textbf{16. If}\space\theta\space\textbf{is the angle between two vectors}\space\\\vec{\textbf{a}}\space\textbf{and}\space\vec{\textbf{b}}\textbf{,}\space\textbf{then\space}\vec{\textbf{a}}.\vec{\textbf{b}}\geq,\space\textbf{only when :}\\\textbf{(a)\space 0}\lt\theta\lt\frac{\pi}{\textbf{2}}\\\textbf{(b)\space} \textbf{0} \leq\theta\leq\frac{\pi}{\textbf{2}}\\\textbf{(c)\space 0}\lt\theta\lt\pi\\\textbf{(d)\space} \textbf{0}\leq\theta\leq\pi\\\textbf{Sol.\space}(b)\space 0\leq\theta\leq\frac{\pi}{2}\\\text{It is given that}\space\vec{a}.\vec{b}\geq 0$$

We know that

$$\vec{a}.\vec{b} = |\vec{a}||\vec{b}|\space\text{cos}\space\theta\geq0\\\Rarr\space\text{cos}\theta\geq 0,\\\lbrack|\vec{a}|\text{and}\space|\vec{b}|\space\text{are positive}\rbrack\\\Rarr\space 0\leq\theta\leq\frac{\pi}{2}$$

$$\textbf{17.\space Let}\space\vec{\textbf{a}}\space\textbf{and}\space\vec{\textbf{b}}\space\textbf{two unit vectors and}\space\theta\\\textbf{is the angle between them. Then,}\\\vec{\textbf{a}} \textbf{+}\vec{\textbf{b}}\space\textbf{is a unit vector, if :}\\\textbf{(a)\space}\theta = \frac{\pi}{\textbf{4}}\\\textbf{(b)\space}\theta = \frac{\pi}{\textbf{3}}\\\textbf{(c)\space}\theta = \frac{\pi}{\textbf{2}}\\\textbf{(d)\space}\theta = \frac{\textbf{2}\pi}{\textbf{3}}\\\textbf{Sol.\space}\theta =\frac{2\pi}{3}\\\text{Let}\space\vec{a}\space\text{and}\space\vec{b}\space\text{be two unit vectors and}\space\theta\\\text{be the angle between them.}\\\text{Then,}\space|\vec{a}| = |\vec{b}|=1$$

$$\text{Now,\space}(\vec{a} +\vec{b})\space\text{is a unit vector if}\\|\vec{a} +\vec{b}|=1\\\Rarr\space (\vec{a} + \vec{b})^{2}=1\\\Rarr\space (\vec{a} + \vec{b}).(\vec{a} +\vec{b})=1\\\Rarr\space \vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{b}.\vec{a} +\vec{b}.\vec{b}= 1\\\Rarr\space|\vec{a}|^{2} + |\vec{b}|^{2} + 2\vec{a}.\vec{b}=1\\\lbrack\because\space \vec{a}.\vec{b} =\vec{b}.\vec{a}\rbrack\\\Rarr\space 1^{2}+1^{2} + 2\vec{a}.\vec{b}=1\\\Rarr\space\vec{a}.\vec{b}=-\frac{1}{2}\\\Rarr\space|\vec{a}||\vec{b}|\text{cos}\space\theta =-\frac{1}{2}\\\Rarr\space 1×1×\text{cos}\space\theta =-\frac{1}{2}\\\Rarr\space \text{cos}\space\theta =-\frac{1}{2}$$

$$\Rarr\space \theta =\frac{2\pi}{3}$$

$$\textbf{18. The value of}\space\hat{\textbf{i}.}\textbf{(}\hat{\textbf{j}}×\hat{\textbf{k}})\textbf{+} \hat{\textbf{j}}.(\hat{\textbf{k}}\textbf{×}\hat{\textbf{i}}) \textbf{+}\hat{\textbf{k}}\textbf{.}\\\textbf{(}\hat{\textbf{i}}×\hat{\textbf{j}})\space\textbf{is}\textbf{:}$$

(a) 0

(b) – 1

(c) 1

(d) 3

Sol. (d) 3

We have,

$$\hat{i}.(\hat{j}×\hat{k}) + \hat{j}.(\hat{k}×\hat{i}) + \hat{k}.(\hat{i}×\hat{j})\\=\hat{i}.\hat{i}+\hat{j}.\hat{j}+\hat{k}.\hat{k}\\\lbrack\because\space\hat{j}×\hat{k} =\hat{i},\hat{k}×\hat{i}=\hat{j},\hat{i}×\hat{j}=\hat{k}\rbrack$$

= 1 + 1 + 1 = 3

$$\lbrack\because\space\hat{i}.\hat{i} =\hat{j}.\hat{j} =\hat{k}.\hat{k}=1\rbrack$$

$$\textbf{19.\space If}\space\theta\space\textbf{is the angle between any two vectors}\\\vec{\textbf{a}}\space\textbf{and}\space\vec{\textbf{b}}\space\textbf{then}\textbf{|}\vec{\textbf{a}} .\vec{\textbf{b}}\textbf{|} \textbf{=} \textbf{|}\vec{\textbf{a}}×\vec{\textbf{b}}|\\\textbf{when}\space\theta\space\textbf{is equal to :}\\\textbf{(a)}\space\textbf{0}\\\textbf{(b)\space}\frac{\pi}{\textbf{4}}\\\textbf{(c)\space}\frac{\pi}{\textbf{2}}\\\textbf{(d)\space}\pi\\\textbf{Sol.\space}\frac{\pi}{4}$$

We have

$$|\vec{a}.\vec{b}| =|\vec{a}×\vec{b}|\\\therefore\space|\vec{a}||\vec{b}|\space\text{cos}\space\theta = |\vec{a}||\vec{b}|\text{sin}\space\theta\\\Rarr\space\text{cos}\space\theta = \text{sin}\space\theta\\\lbrack\because\space|\vec{a}|\space\text{and}\space|\vec{b}|\space\text{are positive}\rbrack\\\Rarr\space\text{tan}\space\theta=1\\\Rarr\space\theta = \frac{\pi}{4}$$

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