Q74 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
If $\bar{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\bar{b} = \hat{i} + 3\hat{k}$ and $\bar{c}$ is a unit vector, then the maximum value of the scalar triple product $[\bar{a}\ \bar{b}\ \bar{c}]$ is
- A$\sqrt{10} + \sqrt{6}$
- B$\sqrt{10}$
- C$\sqrt{6}$
- Correct answerD$\sqrt{59}$
Solution
The scalar triple product is given by $[\bar{a}\ \bar{b}\ \bar{c}] = (\bar{a} \times \bar{b}) \cdot \bar{c}$.
First, calculating the cross product $\bar{a} \times \bar{b}$:
$\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix}$
$= \hat{i}(3 - 0) - \hat{j}(6 - (-1)) + \hat{k}(0 - 1)$
$= 3\hat{i} - 7\hat{j} - \hat{k}$
The scalar triple product becomes $(3\hat{i} - 7\hat{j} - \hat{k}) \cdot \bar{c}$.
Using the dot product formula, $(\bar{a} \times \bar{b}) \cdot \bar{c} = |\bar{a} \times \bar{b}| |\bar{c}| \cos \theta$, where $\theta$ is the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$.
Since $\bar{c}$ is a unit vector, $|\bar{c}| = 1$. The maximum value occurs when $\cos \theta = 1$ (i.e., when $\bar{c}$ is parallel to $\bar{a} \times \bar{b}$).
Maximum value $= |\bar{a} \times \bar{b}|$
$= \sqrt{3^2 + (-7)^2 + (-1)^2}$
$= \sqrt{9 + 49 + 1} = \sqrt{59}$
Answer: $\sqrt{59}$
First, calculating the cross product $\bar{a} \times \bar{b}$:
$\bar{a} \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix}$
$= \hat{i}(3 - 0) - \hat{j}(6 - (-1)) + \hat{k}(0 - 1)$
$= 3\hat{i} - 7\hat{j} - \hat{k}$
The scalar triple product becomes $(3\hat{i} - 7\hat{j} - \hat{k}) \cdot \bar{c}$.
Using the dot product formula, $(\bar{a} \times \bar{b}) \cdot \bar{c} = |\bar{a} \times \bar{b}| |\bar{c}| \cos \theta$, where $\theta$ is the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$.
Since $\bar{c}$ is a unit vector, $|\bar{c}| = 1$. The maximum value occurs when $\cos \theta = 1$ (i.e., when $\bar{c}$ is parallel to $\bar{a} \times \bar{b}$).
Maximum value $= |\bar{a} \times \bar{b}|$
$= \sqrt{3^2 + (-7)^2 + (-1)^2}$
$= \sqrt{9 + 49 + 1} = \sqrt{59}$
Answer: $\sqrt{59}$
Hence, the correct answer is option: 'D'.