Vectors 5 Question 14
14. If $\overrightarrow{\mathbf{A}}=2 \hat{\mathbf{i}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{B}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{C}}=4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}$. Determine a vector $\overrightarrow{\mathbf{R}}$ satisfying $\overrightarrow{\mathbf{R}} \times \overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{C}} \times \overrightarrow{\mathbf{B}}$ and $\overrightarrow{\mathbf{R}} \cdot \overrightarrow{\mathbf{A}}=0$.
(1990, 3M)
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Solution:
- Let $\overrightarrow{\mathbf{R}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
$$ \begin{aligned} & \therefore \quad \overrightarrow{\mathbf{R}} \times \overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{C}} \times \overrightarrow{\mathbf{B}} \\ & \Rightarrow \quad\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ x & y & z \\ 1 & 1 & 1 \end{array}\right|=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & -3 & 7 \\ 1 & 1 & 1 \end{array}\right| \\ & \Rightarrow \quad(y-z) \hat{\mathbf{i}}-(x-z) \hat{\mathbf{j}}+(x-y) \hat{\mathbf{k}}=-10 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}} \\ & \Rightarrow y-z=-10, z-x=-3, x-y=7 \\ & \text { and } \quad \overrightarrow{\mathbf{R}} \cdot \overrightarrow{\mathbf{A}}=0 \Rightarrow 2 x+z=0 \end{aligned} $$
On solving above equations, $x=-1, y=-8$ and $z=2$
$$ \therefore \quad \overrightarrow{\mathbf{R}}=-\hat{\mathbf{i}}-8 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} $$
