Question

The magnitude of projection of line joining (3, 4, 5) and (4, 6, 3) on the line joining (−1, 2, 4) and (1, 0, 5) is

• A. \( \frac{4}{3} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{8}{3} \)
• D. \( \frac{1}{3} \)

Hint:

Remember that "projection" can be a scalar value (which can be negative), but "magnitude of projection" or "length of projection" is always the absolute value.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The projection of a vector $\vec{AB}$ on another vector $\vec{CD}$ is given by the formula: \[ \text{Projection} = \frac{|\vec{AB} \cdot \vec{CD}|}{|\vec{CD}|} \]

Step 2: Calculating the Vectors
1. Let the first line segment be $AB$. $\vec{AB} = (4-3)\hat{i} + (6-4)\hat{j} + (3-5)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k}$ 2. Let the second line segment (on which we project) be $CD$. $\vec{CD} = (1 - (-1))\hat{i} + (0-2)\hat{j} + (5-4)\hat{k} = 2\hat{i} - 2\hat{j} + \hat{k}$

Step 3: Applying the Projection Formula
1. Calculate the dot product $\vec{AB} \cdot \vec{CD}$: $(1)(2) + (2)(-2) + (-2)(1) = 2 - 4 - 2 = -4$ 2. Calculate the magnitude $|\vec{CD}|$: $\sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$ 3. Magnitude of Projection: $\left| \frac{-4}{3} \right| = \frac{4}{3}$

Step 4: Final Answer
The magnitude of the projection is \( \frac{4}{3} \).