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

The Correct Option is D

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.