Vectors 4 Question 5
5. The unit vector which is orthogonal to the vector $3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and is coplanar with the vectors $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is
(2004, 1M)
(a) $\frac{2 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{41}}$
(b) $\frac{2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}}{\sqrt{\mathbf{1 3}}}$
(c) $\frac{3 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{10}}$
(d) $\frac{4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}}{\sqrt{34}}$
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Answer:
Correct Answer: 5. (c)
Solution:
- As we know that, a vector coplanar to $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and orthogonal to $\overrightarrow{\mathbf{c}}$ is $\lambda{(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}}$.
$\therefore$ A vector coplanar to $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}),(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$ and orthogonal to $3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$
$$ \begin{aligned} & =\lambda[{(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})} \times(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})] \\ & =\lambda[(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \times(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})] \\ & =\lambda(21 \hat{\mathbf{j}}-7 \hat{\mathbf{k}}) \end{aligned} $$
$\therefore$ Unit vector $=+\frac{(21 \hat{\mathbf{j}}-7 \hat{\mathbf{k}})}{\sqrt{(21)^{2}+(7)^{2}}}=+\frac{(3 \hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{10}}$
