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.