Question

The vector(s) which are coplanar with vectors $\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + 2\mathop k\limits^ \wedge$ and $\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge + \mathop k\limits^ \wedge$, are perpendicular to the vector $\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge$ are:
A $\mathop j\limits^ \wedge - \mathop k\limits^ \wedge$
B $- \mathop i\limits^ \wedge + \mathop j\limits^ \wedge$
C $\mathop i\limits^ \wedge - \mathop j\limits^ \wedge$
D $- \mathop j\limits^ \wedge + \mathop k\limits^ \wedge$

Answer 1

Coplanar vectors are the vectors which lie on the same plane, in a three-dimensional space. These are vectors which are parallel to the same plane. We can always find in a plane any two random vectors, which are coplanar, hence we can find by solving for the vectors which are coplanar with respect to the perpendicular vector.

Complete step-by-step solution:
Let us write the given vectors i.e.,
$\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + 2\mathop k\limits^ \wedge$ and $\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge + \mathop k\limits^ \wedge$
And the given vectors are perpendicular to the vector $\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge$.
Now let $\mathop a\limits^ \wedge$= $\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + 2\mathop k\limits^ \wedge$, $\mathop b\limits^ \wedge$= $\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge + \mathop k\limits^ \wedge$ and $\mathop c\limits^ \wedge$= $\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge$
Let the vector on the plane of $\mathop a\limits^ \wedge$ and $\mathop b\limits^ \wedge$ is:
$\mathop r\limits^ \wedge = \lambda \mathop a\limits^ \wedge + \mu \mathop b\limits^ \wedge$
Now, substitute the vectors of a and b as
$\mathop r\limits^ \wedge = \lambda \left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + 2\mathop k\limits^ \wedge } \right) + \mu \left( {\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right)$
$\mathop r\limits^ \wedge = \left( {\lambda + \mu } \right)i + \left( {\lambda + 2\mu } \right)j + \left( {2\lambda + \mu } \right)k$
Also,
$\mathop r\limits^ \wedge \cdot \mathop c\limits^ \wedge = 0$
\Rightarrow $$$$\left( {\lambda + \mu } \right) \cdot 1 + \left( {\lambda + 2\mu } \right) \cdot 1 + \left( {2\lambda + \mu } \right) \cdot 1 = 0
\Rightarrow $$$$4\lambda + 4\mu = 0
\Rightarrow $$$$\lambda + \mu = 0
Hence,
$\left[ {\mathop r\limits^ \wedge \mathop a\limits^ \wedge \mathop b\limits^ \wedge } \right] = 0$
Therefore, vectors $\mathop i\limits^ \wedge - \mathop j\limits^ \wedge$ and $- \mathop j\limits^ \wedge + \mathop k\limits^ \wedge$ satisfies the given condition.

Hence, the answer is both option C and D.

Note: In this above question, we drew a figure of a triangle. Drawing a figure really helped to solve the question. While solving any trigonometric ratio related question, it is highly recommended to draw a diagram which will not only help in clearing the confusion but will also help in easily solving the question.