Q120 / 150 · MHT CET 2026 (15 April Shift 1) · Physics
A vector $\vec{A}$ when added to the sum of the vectors $(\hat{i} - 2\hat{j} + 2\hat{k})$ and $(-2\hat{i} + \hat{j} - \hat{k})$ gives a unit vector along Y axis. The magnitude of vector $\vec{A}$ is
- A$\sqrt{3}$
- Correct answerB$\sqrt{6}$
- C$\sqrt{8}$
- D$\sqrt{10}$
Solution
Let the given vectors be $\vec{V}_1 = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{V}_2 = -2\hat{i} + \hat{j} - \hat{k}$.
The sum of these vectors is $\vec{V}_1 + \vec{V}_2 = -\hat{i} - \hat{j} + \hat{k}$.
According to the given condition, $\vec{A} + (\vec{V}_1 + \vec{V}_2) = \hat{j}$.
Substituting the sum, we get $\vec{A} + (-\hat{i} - \hat{j} + \hat{k}) = \hat{j}$.
$\vec{A} = \hat{j} - (-\hat{i} - \hat{j} + \hat{k}) = \hat{i} + 2\hat{j} - \hat{k}$.
The magnitude of vector $\vec{A}$ is $|\vec{A}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}$.
Answer: $\sqrt{6}$
The sum of these vectors is $\vec{V}_1 + \vec{V}_2 = -\hat{i} - \hat{j} + \hat{k}$.
According to the given condition, $\vec{A} + (\vec{V}_1 + \vec{V}_2) = \hat{j}$.
Substituting the sum, we get $\vec{A} + (-\hat{i} - \hat{j} + \hat{k}) = \hat{j}$.
$\vec{A} = \hat{j} - (-\hat{i} - \hat{j} + \hat{k}) = \hat{i} + 2\hat{j} - \hat{k}$.
The magnitude of vector $\vec{A}$ is $|\vec{A}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}$.
Answer: $\sqrt{6}$
Hence, the correct answer is option: 'B'.