Q70 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
Let $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = |\vec{c}|$, $|\vec{c} - \vec{a}| = 2\sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^\circ$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to
- A$\dfrac{3\sqrt{3}}{2}$
- B$\sqrt{\dfrac{3}{2}}$
- Correct answerC$3\sqrt{\dfrac{3}{2}}$
- D$\dfrac{9\sqrt{3}}{2}$
Solution
Given $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
The magnitude of $\vec{a}$ is $|\vec{a}| = \sqrt{1^2 + 2^2 + (-2)^2} = 3$.
Squaring the given equation $|\vec{c} - \vec{a}| = 2\sqrt{2}$ yields:
$|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c} = 8$
Substituting $|\vec{a}| = 3$ and $\vec{a} \cdot \vec{c} = |\vec{c}|$:
$|\vec{c}|^2 + 9 - 2|\vec{c}| = 8$
$|\vec{c}|^2 - 2|\vec{c}| + 1 = 0$
$(|\vec{c}| - 1)^2 = 0 \Rightarrow |\vec{c}| = 1$
Calculating the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = (\hat{i} + 2\hat{j} - 2\hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) = -3\hat{j} - 3\hat{k}$
The magnitude is $|\vec{a} \times \vec{b}| = \sqrt{(-3)^2 + (-3)^2} = 3\sqrt{2}$.
The angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^\circ$.
Using the definition of the cross product magnitude:
$|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| \sin 60^\circ$
$|(\vec{a} \times \vec{b}) \times \vec{c}| = (3\sqrt{2})(1)\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{3\sqrt{6}}{2} = 3\sqrt{\dfrac{3}{2}}$
Answer: $3\sqrt{\dfrac{3}{2}}$
The magnitude of $\vec{a}$ is $|\vec{a}| = \sqrt{1^2 + 2^2 + (-2)^2} = 3$.
Squaring the given equation $|\vec{c} - \vec{a}| = 2\sqrt{2}$ yields:
$|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c} = 8$
Substituting $|\vec{a}| = 3$ and $\vec{a} \cdot \vec{c} = |\vec{c}|$:
$|\vec{c}|^2 + 9 - 2|\vec{c}| = 8$
$|\vec{c}|^2 - 2|\vec{c}| + 1 = 0$
$(|\vec{c}| - 1)^2 = 0 \Rightarrow |\vec{c}| = 1$
Calculating the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = (\hat{i} + 2\hat{j} - 2\hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) = -3\hat{j} - 3\hat{k}$
The magnitude is $|\vec{a} \times \vec{b}| = \sqrt{(-3)^2 + (-3)^2} = 3\sqrt{2}$.
The angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^\circ$.
Using the definition of the cross product magnitude:
$|(\vec{a} \times \vec{b}) \times \vec{c}| = |\vec{a} \times \vec{b}| |\vec{c}| \sin 60^\circ$
$|(\vec{a} \times \vec{b}) \times \vec{c}| = (3\sqrt{2})(1)\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{3\sqrt{6}}{2} = 3\sqrt{\dfrac{3}{2}}$
Answer: $3\sqrt{\dfrac{3}{2}}$
Hence, the correct answer is option: 'C'.