Question

The vector(s) which is/are coplanar with vectors $\hat{i}+\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$, are perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$ is/are

• A. $\hat{j}-\hat{k}$
• B. $-\hat{i}+\hat{j}$
• C. $\hat{i}-\hat{j}$
• D. $-\hat{j}+\hat{k}$

Answer 1

Answer: (A), (D)

Let $\overrightarrow{a}=\hat{i}+\hat{j}+2\hat{k},\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{c}=\hat{i}+\hat{j}+\hat{k}$
$\therefore$ A vector coplanar to $\overrightarrow{a}$ and $\overrightarrow{b}$, and perpendicular to $\overrightarrow{c}$
$=\lambda (\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}=\lambda \{(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{v}-(\overrightarrow{b}\cdot \overrightarrow{c}\}\overrightarrow{a}\}$
$=\lambda \{(1+1+4)(\hat{i}+2\hat{j}+\hat{k})-(1+2+1)(\hat{i}+\hat{j}+2\hat{k})\}$
$=\lambda \{6\hat{i}+12\hat{j}+6\hat{k}-6\hat{i}-6\hat{j}-12\hat{k}\}$
$=\lambda \{6\hat{j}-6\hat{k}\}=6\lambda \{\hat{j}-\hat{k}\}$
For $\lambda =\frac{1}{6}\Rightarrow$ Option (a) is correct.
and for $\lambda =-\frac{1}{6}\Rightarrow$ Option (d) is correct.