Q74 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
If $\vec{u} = \hat{i} + 2\hat{j} - 2\hat{k}, \vec{v} = 2\hat{i} + \hat{k}$ and $\vec{w}$ is unit vector then the maximum value of scalar triple product $[\vec{u}\ \vec{v}\ \vec{w}]$ is
- A$-3\sqrt{5}$
- B$0$
- Correct answerC$3\sqrt{5}$
- D$\sqrt{54}$
Solution
The scalar triple product is given by $[\vec{u}\ \vec{v}\ \vec{w}] = (\vec{u} \times \vec{v}) \cdot \vec{w}$.
First, we find the cross product $\vec{u} \times \vec{v}$:
$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -2 \\ 2 & 0 & 1 \end{vmatrix}$
$\vec{u} \times \vec{v} = \hat{i}(2 - 0) - \hat{j}(1 - (-4)) + \hat{k}(0 - 4)$
$\vec{u} \times \vec{v} = 2\hat{i} - 5\hat{j} - 4\hat{k}$
The magnitude of this vector is:
$|\vec{u} \times \vec{v}| = \sqrt{2^2 + (-5)^2 + (-4)^2} = \sqrt{4 + 25 + 16} = \sqrt{45} = 3\sqrt{5}$
The dot product $(\vec{u} \times \vec{v}) \cdot \vec{w}$ is maximized when the unit vector $\vec{w}$ is in the same direction as $\vec{u} \times \vec{v}$.
Maximum value $= |\vec{u} \times \vec{v}| |\vec{w}| \cos(0^{\circ})$
Since $|\vec{w}| = 1$, the maximum value is $3\sqrt{5} \times 1 \times 1 = 3\sqrt{5}$.
Answer: $3\sqrt{5}$
First, we find the cross product $\vec{u} \times \vec{v}$:
$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -2 \\ 2 & 0 & 1 \end{vmatrix}$
$\vec{u} \times \vec{v} = \hat{i}(2 - 0) - \hat{j}(1 - (-4)) + \hat{k}(0 - 4)$
$\vec{u} \times \vec{v} = 2\hat{i} - 5\hat{j} - 4\hat{k}$
The magnitude of this vector is:
$|\vec{u} \times \vec{v}| = \sqrt{2^2 + (-5)^2 + (-4)^2} = \sqrt{4 + 25 + 16} = \sqrt{45} = 3\sqrt{5}$
The dot product $(\vec{u} \times \vec{v}) \cdot \vec{w}$ is maximized when the unit vector $\vec{w}$ is in the same direction as $\vec{u} \times \vec{v}$.
Maximum value $= |\vec{u} \times \vec{v}| |\vec{w}| \cos(0^{\circ})$
Since $|\vec{w}| = 1$, the maximum value is $3\sqrt{5} \times 1 \times 1 = 3\sqrt{5}$.
Answer: $3\sqrt{5}$
Hence, the correct answer is option: 'C'.