Question

The straight line passing through the points $(3,2,3)$ and $(5,-1,-2)$ is perpendicular to the straight line passing through the points $(1,3,1)$ and $(\alpha, \alpha, \alpha)$. Then the value of $\alpha$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Hint:

In 3D geometry, "perpendicular" always translates to "Dot Product = 0". Use this immediately after finding direction ratios.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
If two straight lines are perpendicular, the dot product of their direction vectors must be zero.
Step 2: Key Formula or Approach:
Find the direction vector \(\vec{d_1}\) of the first line using \((x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}\).
Find the direction vector \(\vec{d_2}\) of the second line similarly.
Set \(\vec{d_1} \cdot \vec{d_2} = 0\) and solve for \(\alpha\).
Step 3: Detailed Explanation:
Calculate direction vector \(\vec{d_1}\) for the first line:
\[ \vec{d_1} = (5 - 3)\hat{i} + (-1 - 2)\hat{j} + (-2 - 3)\hat{k} = 2\hat{i} - 3\hat{j} - 5\hat{k} \] Calculate direction vector \(\vec{d_2}\) for the second line:
\[ \vec{d_2} = (\alpha - 1)\hat{i} + (\alpha - 3)\hat{j} + (\alpha - 1)\hat{k} \] Since the lines are perpendicular, \(\vec{d_1} \cdot \vec{d_2} = 0\):
\[ 2(\alpha - 1) + (-3)(\alpha - 3) + (-5)(\alpha - 1) = 0 \] Expand and solve:
\[ 2\alpha - 2 - 3\alpha + 9 - 5\alpha + 5 = 0 \] Combine like terms:
\[ (2 - 3 - 5)\alpha + (-2 + 9 + 5) = 0 \] \[ -6\alpha + 12 = 0 \] \[ 6\alpha = 12 \implies \alpha = 2 \] Step 4: Final Answer:
The value of \(\alpha\) is 2.