Question

If \(\vec a\) is a vector perpendicular to the plane containing the vectors \[ 2\hat i-\hat j+3\hat k \quad\text{and}\quad -\hat i+3\hat j+2\hat k, \] then the magnitude of the projection of the vector \[ 3\hat i+2\hat j-\hat k \] on \(\vec a\) is

• A. \(\dfrac{24}{\sqrt{195}}\)
• B. \(\dfrac{42}{\sqrt{195}}\)
• C. \(2\sqrt{\dfrac{13}{15}}\)
• D. \(4\sqrt{\dfrac{13}{15}}\)

Hint:

If a vector is perpendicular to the plane containing two vectors \(\vec u\) and \(\vec v\), then \[ \boxed{\vec a\parallel(\vec u\times\vec v)}. \] After finding the normal vector, use \[ \boxed{\text{Projection}=\frac{|\vec x\cdot\vec a|}{|\vec a|}}. \]

Answer 1

The Correct Option is D

Step 1: Find a vector perpendicular to the plane.

Let& nbsp;

\[ \vec u=(2,-1,3), \qquad \vec v=(-1,3,2). \]

Since \(\vec a\) is perpendicular to the plane,

\[ \vec a\parallel(\vec u\times\vec v). \]

Now,

\[ \vec u\times\vec v = \begin{vmatrix} \hat{i} & amp; \hat{j} & amp; \hat{k}\\ 2 & amp; -1 & amp; 3\\ -1 & amp; 3 & amp; 2 \end{vmatrix} = (-11,-7,5). \]

Hence, we may take

\[ \vec a=(-11,-7,5). \]

Step 2: Compute the projection.

Let

\[ \vec w=(3,2,-1). \]

Then,

\[ \vec w\cdot\vec a = 3(-11)+2(-7)+(-1)(5) = -52. \]

Therefore,

\[ |\vec w\cdot\vec a| = 52. \]

Also,

\[ |\vec a| = \sqrt{121+49+25} = \sqrt{195}. \]

Hence, the magnitude of the projection is

\[ \frac{52}{\sqrt{195}}. \]

Simplifying,

\[ \frac{52}{\sqrt{195}} = 4\sqrt{\frac{13}{15}}. \]

Step 3: Final conclusion.

\[ \boxed{4\sqrt{\frac{13}{15}}} \]

Hence, the correct option is \[ \boxed{(D)}. \]