Q71 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics

A vector which is orthogonal to the vector $\bar{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and coplanar with the vectors $\bar{b} = 3\hat{i} + 2\hat{j}$ and $\bar{c} = 2\hat{i} + \hat{j} + 3\hat{k}$ is

  • Correct answer
    A
    $25\hat{i} + 19\hat{j} - 21\hat{k}$
  • B
    $-25\hat{i} + 19\hat{j} - 21\hat{k}$
  • C
    $-25\hat{i} + 19\hat{j} + 21\hat{k}$
  • D
    $25\hat{i} + 19\hat{j} + 21\hat{k}$

Solution

Let the required vector be $\bar{v}$.

Since $\bar{v}$ is coplanar with $\bar{b}$ and $\bar{c}$, and orthogonal to $\bar{a}$, it is parallel to the vector triple product $\bar{a} \times (\bar{b} \times \bar{c})$.

Using the vector triple product expansion:

$\bar{v} = \bar{a} \times (\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c})\bar{b} - (\bar{a} \cdot \bar{b})\bar{c}$

Given vectors are:

$\bar{a} = \hat{i} + 2\hat{j} + 3\hat{k}$

$\bar{b} = 3\hat{i} + 2\hat{j}$

$\bar{c} = 2\hat{i} + \hat{j} + 3\hat{k}$

Calculating the dot products:

$\bar{a} \cdot \bar{c} = (1)(2) + (2)(1) + (3)(3) = 2 + 2 + 9 = 13$

$\bar{a} \cdot \bar{b} = (1)(3) + (2)(2) + (3)(0) = 3 + 4 + 0 = 7$

Substituting these values into the expansion:

$\bar{v} = 13(3\hat{i} + 2\hat{j}) - 7(2\hat{i} + \hat{j} + 3\hat{k})$

$\bar{v} = (39\hat{i} + 26\hat{j}) - (14\hat{i} + 7\hat{j} + 21\hat{k})$

$\bar{v} = (39 - 14)\hat{i} + (26 - 7)\hat{j} - 21\hat{k}$

$\bar{v} = 25\hat{i} + 19\hat{j} - 21\hat{k}$

Answer: $25\hat{i} + 19\hat{j} - 21\hat{k}$

Hence, the correct answer is option: 'A'.

Relevant Questions