Question

Let \(\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k}\) and \(\vec{b} = 2\hat{i} + \hat{j} + \alpha\hat{k}\). If \(|\vec{a} + \vec{b}| = \sqrt{29}\). Then the possible values of \(\alpha\) are

• A. \(\pm 2\)
• B. \(\pm 3\)
• C. 3, 7
• D. \(\pm 5\)
• E. -3, -7

Hint:

When you have a squared term like \((x-c)^2 = k\), always consider both the positive and negative roots. This usually leads to two possible solutions in vector coordinate problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We first add the two vectors component-wise and then use the magnitude formula to set up an equation for \(\alpha\).

Step 2: Key Formula or Approach:
1. Addition: \(\vec{a} + \vec{b} = (a_x+b_x)\hat{i} + (a_y+b_y)\hat{j} + (a_z+b_z)\hat{k}\).
2. Magnitude: \(|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}\).

Step 3: Detailed Explanation:

Step 1: Find \(\vec{a} + \vec{b}\).
\[ \vec{a} + \vec{b} = (2+2)\hat{i} + (2+1)\hat{j} + (-5+\alpha)\hat{k} = 4\hat{i} + 3\hat{j} + (\alpha - 5)\hat{k} \]

Step 2: Use the magnitude condition.
\[ |\vec{a} + \vec{b}|^2 = (\sqrt{29})^2 \]
\[ 4^2 + 3^2 + (\alpha - 5)^2 = 29 \]
\[ 16 + 9 + (\alpha - 5)^2 = 29 \]
\[ 25 + (\alpha - 5)^2 = 29 \]
\[ (\alpha - 5)^2 = 4 \]

Step 3: Solve for \(\alpha\).
\[ \alpha - 5 = 2 \implies \alpha = 7 \]
\[ \alpha - 5 = -2 \implies \alpha = 3 \]

Step 4: Final Answer:
The possible values of \(\alpha\) are 3 and 7.