Vectors 4 Question 13
13. If $\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}}$ and $\overrightarrow{\mathbf{C}}$ are vectors such that $|\overrightarrow{\mathbf{B}}|=|\overrightarrow{\mathbf{C}}|$. Prove that $[(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}) \times(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{C}})] \times(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}}) \cdot(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}})=\overrightarrow{\mathbf{0}}$.
(1997, 5M)
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Answer:
Correct Answer: 13. (i) $\overrightarrow{\mathbf{X}}=\left(\frac{\overrightarrow{\mathbf{c}}}{|\overrightarrow{\mathbf{A}}|^{2}}\right) \overrightarrow{\mathbf{A}}-\left(\frac{1}{|\overrightarrow{\mathbf{A}}|^{2}}\right)(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}) \quad$ (ii) $\left(A _2 \hat{\mathbf{i}}-A _1 \hat{\mathbf{j}}+A _3 \hat{\mathbf{k}}\right)$
Solution:
- Now, $(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}) \times(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{C}})$
$$ \begin{aligned} = & \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}} \\ = & \overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}} \quad[\because \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{A}}=0] \\ \therefore \quad[(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}) \times & (\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{C}})] \times(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}}) \\ = & {[\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}}] \times(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}}) } \\ = & (\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}) \times(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}})+(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{C}}) \times(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}}) \\ = & {(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}) \cdot \overrightarrow{\mathbf{C}}} \overrightarrow{\mathbf{B}}-{(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}) \cdot \overrightarrow{\mathbf{B}}} \overrightarrow{\mathbf{C}} \\ & \quad+{(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{C}}) \cdot \overrightarrow{\mathbf{C}}} \overrightarrow{\mathbf{B}}-{(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{C}}) \cdot \overrightarrow{\mathbf{B}}} \overrightarrow{\mathbf{C}} \\ & \quad[\because(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{a}}] \\ = & {[\overrightarrow{\mathbf{B}} \overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{C}}] \overrightarrow{\mathbf{B}}-[\overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{C}} \overrightarrow{\mathbf{B}}] \overrightarrow{\mathbf{C}} } \end{aligned} $$
$[\because[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=0$, if any two of $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are equal ] = [$\overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{C}} \overrightarrow{\mathbf{B}}$] $(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{C}})$
Now, $\quad[(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}) \times(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{C}})] \times(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}}) \cdot(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}})$
$$ \begin{aligned} & =([\overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{C}} \overrightarrow{\mathbf{B}}]{\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{C}}}) \cdot(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}) \\ & =[\overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{C}} \overrightarrow{\mathbf{B}}]{(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{C}}) \cdot(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}})} \\ & =[\overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{C}} \overrightarrow{\mathbf{B}}] {|\overrightarrow{\mathbf{B}}|^{2}-|\overrightarrow{\mathbf{C}}|^{2}} =\overrightarrow{0} \quad[\because|\overrightarrow{\mathbf{B}}|=|\overrightarrow{\mathbf{C}}|, \text { given }] \end{aligned} $$
