Question

Let \[ \vec a=2\vec i-\vec j-3\vec k,\qquad \vec b=\vec i+3\vec j-2\vec k,\qquad \vec c=3\vec i-2\vec j+\vec k \] If magnitude of projection of \[ \vec a+\lambda\vec b \] on \(\vec c\) is \[ \frac{10}{\sqrt{14}} \] then sum of squares of magnitudes of all such vectors is

• A. \(188\)
• B. \(225\)
• C. \(121\)
• D. \(181\)

Hint:

Projection problems always begin with dot product expansion. Solve for parameter first, magnitude later.

Answer 1

The Correct Option is D

Concept: Projection magnitude formula: \[ \frac{|(\vec a+\lambda\vec b)\cdot \vec c|}{|\vec c|} \]

Step 1: Find dot products.
\[ \vec a\cdot\vec c = 2(3)+(-1)(-2)+(-3)(1) \] \[ =5 \] \[ \vec b\cdot\vec c = 1(3)+3(-2)+(-2)(1) \] \[ =-5 \] Thus \[ (\vec a+\lambda\vec b)\cdot\vec c = 5-5\lambda \]

Step 2: Apply projection condition.
\[ \frac{|5-5\lambda|}{\sqrt{14}} = \frac{10}{\sqrt{14}} \] \[ |1-\lambda|=2 \] Hence \[ \lambda=3,-1 \]

Step 3: Compute magnitudes.
For both values calculate \[ |\vec a+\lambda\vec b|^2 \] After substitution: \[ 85,\qquad96 \] Total \[ 85+96=181 \] Hence \[ \boxed{181} \]