# NCERT Solutions for Class 12 Maths Chapter 10 Vectors

Q. Find the angles of a triangle ABC whose vertices are A(1, 2, 3), B(2, 4, 8) and C(3, 6, 7).

$$\textbf{Ans. } \text{Let O be the origin.}\\ \text{Then, }\space \space\overrightarrow{OA}=\hat{i}+2\hat{j}+3\hat{k},\\ \overrightarrow{OB}=\hat2{i}+4\hat{j}+8\hat{k} \\ \text{and }\space \space \overrightarrow{OC}=\hat3{i}+6\hat{j}+7\hat{k} \\ \therefore \space \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} \\ = \Big(2\hat{i}+4\hat{j}+8\hat{k}\Big)-\Big(\hat{i}+2\hat{j}+3\hat{k}\Big)\\=\hat{i}+2\hat{j}+5\hat{k},\\ \bigm\vert\overrightarrow{AB} \bigm\vert \space = \space \sqrt{ \mathstrut 30},\\ \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}\\ = \Big(3\hat{i}+6\hat{j}+7\hat{k}\Big)-\Big(2\hat{i}+4\hat{j}+8\hat{k}\Big)\\ = \hat{i}+2\hat{j}-\hat{k},\\ \bigm\vert\overrightarrow{BC} \bigm\vert \space = \space \sqrt{ \mathstrut 6}\\ \overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC} \\ = \Big(\hat{i}+2\hat{j}+3\hat{k}\Big)-\Big(3\hat{i}+6\hat{j}+7\hat{k}\Big)\\ = -\space 2\hat{i}-4\hat{j}-4\hat{k},\\\bigm\vert\overrightarrow{CA} \bigm\vert \space = \space \sqrt{ \mathstrut 36}\space =6\\ \text{Now, }\space \overrightarrow{AB}+\overrightarrow{BC}=2\hat{i}+4\hat{j}+4\hat{k}\\ \text{Also, }\space \overrightarrow{AB}+ \overrightarrow{BC}+\space \overrightarrow{CA}=\overrightarrow{0}\\ \text{Which show that A, B and C are the vertices of a triangle.}\\ \text{Now, }\space \overrightarrow{AB} \cdot \overrightarrow{BC}=\Big(\hat{i}+2\hat{j}+5\hat{k} \Big).\Big(\hat{i}+2\hat{j}-\hat{k}\Big)\\ = \space 1+4-5=0 \\ \Rarr \space \space AB\space \bot\space BC\space \text{and therefore,}\space \angle B=90^o\\ \text{Now, }\space \angle A \space \text{is the angle between }\overrightarrow{AB}\space \text{and }\space \overrightarrow{AC}\space \text{and }\space \overrightarrow{AB}.\space \overrightarrow{AC}\space = \Big(\hat{i}+2\hat{j}+5\hat{k}\Big).\Big(-2\hat{i}-4\hat{j}-4\hat{k}\Big)\\ =-2-8-20 \\ =-30 \\ cos \space A=\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{\bigm\vert\overrightarrow{AB} \bigm\vert\space \bigm\vert\overrightarrow{AC} \bigm\vert}\\ = \frac{30}{\sqrt[]{30}×6}=\frac{\sqrt[]{30}}{6}\\ \text{But, }\space \angle C \space \text{is the angle between }\overrightarrow{CA}\space \text{and }\space \overrightarrow{CB}\\ \text{Hence, }\space cos \space C =\frac{\overrightarrow{CA}.\overrightarrow{CB}}{\bigm\vert\overrightarrow{CA} \bigm\vert\space \bigm\vert\overrightarrow{CB} \bigm\vert}\\ =\frac{\Big(-2\hat{i}-4\hat{j}-4\hat{k}). \Big(-\hat{i}-2\hat{j}+\hat{k}\Big)}{6\sqrt[]{6}}\\ =\frac{2+8-4}{6\sqrt[]{6}}=\frac{6}{6\sqrt[]{6}}=\frac{1}{\sqrt[]{6}}=\frac{\sqrt[]{6}}{6} \\ \text{Hence, ABC is right-angled triangle,\space right-angled at B. The other two angles are}\\ cos^{-1}\space \frac{\sqrt[]{30}}{6}\space \text{and }\space cos^{-1}\frac{\sqrt[]{6}}{6}$$

Q. If the diagonals of a parallelogram are determined by the vectors $$\space \overrightarrow{a}\space and \space \overrightarrow{b},\textbf{show that its area is}\frac{1}{2}\space\bigm\vert\overrightarrow{a}×\overrightarrow{b}\bigm\vert.$$

Ans. Let ABCD be a parallelogram. Its diagonals AC and BD intersect at O.

$$\text{Let }\space \space \overrightarrow{AC}=\space \overrightarrow{a}\\ \text{and} \space \space \overrightarrow{BD}=\overrightarrow{b}\\ \text{Since the diaglonals of a parallelogram bisect each other, O is the mid-point of AC and BD.}\\ \therefore \space\space \overrightarrow{OD}=\frac{1}{2}\overrightarrow{BD}=\frac{\overrightarrow{b}}{2},\\\overrightarrow{OC}=\space \frac{1}{2}\overrightarrow{AC}=\frac{\overrightarrow{a}}{2},\\ \overrightarrow{OB}=-\frac{1}{2}\overrightarrow{BD}=- \frac{1}{2}\overrightarrow{b}\\\text{and }\space \overrightarrow{OA}\space =-\space \frac{1}{2}\space \overrightarrow{AC}=-\frac{1}{2}\space \overrightarrow{a}\\ \therefore \space \overrightarrow{AB} \space =\space \overrightarrow{OB}\space -\space \overrightarrow{OA}\\ = \space \Bigg(-\frac{1}{2}\overrightarrow{b}\Bigg)-\Bigg(-\frac{1}{2}\overrightarrow{a}\Bigg)\\ =\space \frac{1}{2}\Big(\overrightarrow{a}-\overrightarrow{b}\Big)\\\text{and} \space \space \overrightarrow{AD}=\space \overrightarrow{OD}\space -\space \overrightarrow{OA}\\ =\space \frac{\overrightarrow{b}}{2}-\Bigg(- \frac{\overrightarrow{a}}{2}\Bigg)\space =\space \frac{1}{2}\Big(\overrightarrow{a}+\overrightarrow{b}\Big)\\ \text{Therefore, the area of parallelogram ABCD}\\ =\space \bigm\vert\overrightarrow{AB}\space ×\space \overrightarrow{AD}\bigm\vert \\ = \space \bigm\vert \frac{1}{2}\Big(\overrightarrow{a}-\overrightarrow{b}\Big)×\space\frac{1}{2} \Big(\overrightarrow{a}+\overrightarrow{b}\Big)\bigm\vert\\ =\space \bigm\vert\frac{1}{4}\Big(\overrightarrow{a}×\overrightarrow{a}+\overrightarrow{a}×\overrightarrow{b}-\overrightarrow{b}×\overrightarrow{a}-\overrightarrow{b}×\overrightarrow{b}\Big)\bigm\vert\\= \space \frac{1}{4} \bigm\vert 0\space +\space \overrightarrow{a}×\overrightarrow{b}+\overrightarrow{a}×\overrightarrow{b}\space -0\bigm\vert\\ =\space \frac{1}{4}\bigm\vert2 \Big(\overrightarrow{a}×\overrightarrow{b}\Big)\bigm\vert\\=\space \frac{1}{2}\bigm\vert\overrightarrow{a}×\overrightarrow{b}\bigm\vert$$