Vectors 4 Question 10
10. The vector(s) which is/are coplanar with vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$, are perpendicular to the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is/are
(2011)
(a) $\hat{\mathbf{j}}-\hat{\mathbf{k}}$
(b) $-\hat{\mathbf{i}}+\hat{\mathbf{j}}$
(c) $\hat{\mathbf{i}}-\hat{\mathbf{j}}$
(d) $-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
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Answer:
Correct Answer: 10. (a, d)
Solution:
- Let $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\therefore$ A vector coplanar to $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, and perpendicular to $\overrightarrow{\mathbf{c}}$ $=\lambda(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=\lambda{(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{v}}-(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{a}}}$
$$ \begin{aligned} & =\lambda{(1+1+4)(\hat{i}+2 \hat{j}+\hat{k})-(1+2+1)(\hat{i}+\hat{j}+2 \hat{k})} \\ & =\lambda{6 \hat{i}+12 \hat{j}+6 \hat{k}-6 \hat{i}-6 \hat{j}-12 \hat{k}} \\ & =\lambda{6 \hat{j}-6 \hat{k}}=6 \lambda{\hat{j}-\hat{k}} \end{aligned} $$
For $\lambda=\frac{1}{6} \Rightarrow$ Option (a) is correct.
and for $\lambda=-\frac{1}{6} \Rightarrow$ Option (d) is correct.
