Question

Let \(\rm \vec a =\hat i +\hat j +\hat k,\; \vec b =\hat i -\hat j + \hat k\) and c = î - ĵ - k̂ be three vectors. A vector \(\rm \vec v\) in the plane of \(\rm \vec a\) and \(\rm \vec b\) whose projection on \(\rm \frac {\vec c} {|\vec c|}\) is \(\frac 1 {\sqrt 3},\) is
1. 3î - ĵ + 3k̂
2. î - 3ĵ + 3k̂
3. 5î - 2ĵ + 5k̂
4. 2î - ĵ + 3k̂

Answer 1

Correct Answer - Option 1 : 3î - ĵ + 3k̂

Calculation:

 \(\rm \vec a =\hat i +\hat j +\hat k,\; \vec b =\hat i -\hat j + \hat k\) and c = î - ĵ - k̂

Given:  vector \(\rm \vec v\) in the plane of \(\rm \vec a\) and \(\rm \vec b\) 

Therefore, \(\rm \vec v = \vec a + λ \vec b\)

⇒ \(\rm \vec v =(\hat i +\hat j +\hat k ) \; + λ (\hat i -\hat j + \hat k)\)

= (1 + λ)î + (1 - λ)ĵ  + (1 + λ)k̂                .... (1)

Projection of \(\rm \vec v\) on \(\rm \frac {\vec c} {|\vec c|}\) = \(\frac 1 {\sqrt 3}\)

⇒ \(\rm \vec v\). \(\rm \frac {\vec c} {|\vec c|}\) = \(\frac 1 {\sqrt 3}\)

⇒ \(\frac {(1 + λ) - (1 - λ) - (1 + λ)}{\sqrt3} = \frac {1}{\sqrt 3}\)

⇒ -(1 - λ) = 1

∴ λ = 2

Now, put the value of λ in equation (1), we get

\(\rm \vec v\) = 3î - ĵ + 3k̂