Question:

Let \(\vec{p} = 2\hat{i} + 3\hat{j} + \hat{k}\) and \(\vec{q} = \hat{i} + 2\hat{j} + \hat{k}\) be two vectors. If a vector \(\vec{r} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}\) is perpendicular to each of the vectors \((\vec{p} + \vec{q})\) and \((\vec{p} - \vec{q})\), and \(|\vec{r}| = \sqrt{3}\), then \(|\alpha| + |\beta| + |\gamma|\) is equal to __________.

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Note that $(\vec{p}+\vec{q}) \times (\vec{p}-\vec{q}) = -2(\vec{p} \times \vec{q})$. You can save time by calculating $\vec{p} \times \vec{q}$ directly. The vector $\vec{r}$ will be parallel to it.
Updated On: Jan 20, 2026
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Correct Answer: 3

Solution and Explanation

Step 1: Understanding the Concept:
A vector that is perpendicular to two non-parallel vectors must be parallel to their cross product.
We find the vectors \( \vec{p} + \vec{q} \) and \( \vec{p} - \vec{q} \), calculate their cross product, and then scale it to satisfy the magnitude condition.
Step 2: Key Formula or Approach:
1. \( \vec{u} \perp \vec{v}, \vec{w} \implies \vec{u} \parallel (\vec{v} \times \vec{w}) \).
2. Cross product: \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}. \]
Step 3: Detailed Explanation:
Calculate the target vectors:
\(\vec{p} + \vec{q} = (2+1)\hat{i} + (3+2)\hat{j} + (1+1)\hat{k} = 3\hat{i} + 5\hat{j} + 2\hat{k}\).
\(\vec{p} - \vec{q} = (2-1)\hat{i} + (3-2)\hat{j} + (1-1)\hat{k} = \hat{i} + \hat{j} + 0\hat{k}\).
Find their cross product:
\[ \vec{V} = (\vec{p} + \vec{q}) \times (\vec{p} - \vec{q}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 2 \\ 1 & 1 & 0 \end{vmatrix} \]
\[ \vec{V} = \hat{i}(0-2) - \hat{j}(0-2) + \hat{k}(3-5) = -2\hat{i} + 2\hat{j} - 2\hat{k} \]
The vector \( \vec{r} \) is parallel to \( \vec{V} \), so \[ \vec{r} = \lambda(-2\hat{i} + 2\hat{j} - 2\hat{k}) \]
Magnitude \( |\vec{r}| = \sqrt{4\lambda^2 + 4\lambda^2 + 4\lambda^2} = \sqrt{12\lambda^2} = 2|\lambda|\sqrt{3} \).
Given \( |\vec{r}| = \sqrt{3} \implies 2|\lambda|\sqrt{3} = \sqrt{3} \implies |\lambda| = \frac{1}{2} \).
Thus, \[ \vec{r} = \pm \frac{1}{2}(-2\hat{i} + 2\hat{j} - 2\hat{k}) = \pm(-\hat{i} + \hat{j} - \hat{k}) \]
The components are \( \alpha = \mp 1, \beta = \pm 1, \gamma = \mp 1 \).
The required sum is \[ |\alpha| + |\beta| + |\gamma| = 1 + 1 + 1 = 3. \]
Step 4: Final Answer:
The value of \( |\alpha| + |\beta| + |\gamma| \) is 3.
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