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

Hint: 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.