Question

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

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

Answer 1

Answer: (D)

$-\widehat{j}+\widehat{k}$

Answer 2

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