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: Find the direction vector of first line
Let the points be \[ A(3,4,5), \quad B(4,6,3) \] So, the vector joining these points is \[ \vec{u}=\overrightarrow{AB} =(4-3)\hat{i}+(6-4)\hat{j}+(3-5)\hat{k} \] \[ \vec{u}=\hat{i}+2\hat{j}-2\hat{k} \] Step 2: Find the direction vector of second line
Let the second line join \[ C(-1,2,4), \quad D(1,0,5) \] Then, \[ \vec{v}=\overrightarrow{CD} =(1-(-1))\hat{i}+(0-2)\hat{j}+(5-4)\hat{k} \] \[ \vec{v}=2\hat{i}-2\hat{j}+\hat{k} \] Step 3: Use projection formula
Magnitude of projection of $\vec{u}$ on $\vec{v}$ is \[ \text{Projection}=\frac{|\vec{u}\cdot \vec{v}|}{|\vec{v}|} \] First, find dot product: \[ \vec{u}\cdot \vec{v} =(1)(2)+(2)(-2)+(-2)(1) \] \[ =2-4-2=-4 \] \[ |\vec{u}\cdot \vec{v}|=4 \] Now magnitude of $\vec{v}$: \[ |\vec{v}|=\sqrt{2^2+(-2)^2+1^2} \] \[ =\sqrt{4+4+1} \] \[ =\sqrt{9}=3 \] Step 4: Final calculation
\[ \text{Projection}=\frac{4}{3} \] \[ \boxed{\frac{4}{3}} \] Hence, the correct option is (A).