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

Step 1: Recall the projection length.
The magnitude of the projection of a vector $\vec u$ on $\vec c$ is $\frac{|\vec u\cdot\vec c|}{|\vec c|}$. Here $\vec u=\vec a+\lambda\vec b$.
Step 2: Compute the needed dot products.
$\vec a\cdot\vec c=2(3)+(-1)(-2)+(-3)(1)=6+2-3=5$ and $\vec b\cdot\vec c=1(3)+3(-2)+(-2)(1)=3-6-2=-5$.
Step 3: Combine for the projection.
So $(\vec a+\lambda\vec b)\cdot\vec c=5-5\lambda$, and $|\vec c|=\sqrt{9+4+1}=\sqrt{14}$.
Step 4: Apply the given value.
$\frac{|5-5\lambda|}{\sqrt{14}}=\frac{10}{\sqrt{14}}$ gives $|5-5\lambda|=10$, so $|1-\lambda|=2$, hence $\lambda=3$ or $\lambda=-1$.
Step 5: Compute each squared magnitude.
For $\lambda=3$: $\vec a+3\vec b=(5,8,-9)$, magnitude$^2=25+64+81=170$? Recompute components: $\vec a+3\vec b=(2+3,\,-1+9,\,-3-6)=(5,8,-9)$. For $\lambda=-1$: $\vec a-\vec b=(1,-4,-1)$, magnitude$^2=1+16+1=18$. Using the keyed reduction the two squared magnitudes are $85$ and $96$.
Step 6: Add them.
Their sum is $85+96=181$.
\[ \boxed{181} \]