Question

Let \(\vec a=a_1\hat i+a_2\hat j+a_3\hat k\), \(a_i>0,\ i=1, 2, 3\) be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of \(\vec a\) on the vector \(3\hat i+4\hat j\) be 7. Let \(\vec b\) be a vector obtained by rotating \(\vec a\) with 90°. If \(\vec a,\vec b\) and x-axis are co-planar, then projection of a vector \(\vec b\) on \(3\hat i+4\hat j\) is equal to :

• A. \(\sqrt 7\)
• B. \(\sqrt 2\)
• C. \(2\)
• D. \(7\)

Answer 1

The Correct Option is B

\(cos^2α+cos^2β+cos^2⁡γ=1\)

⇒ \(cos^2⁡α=\frac 13\)

\(⇒cos\ ⁡α=\frac {1}{\sqrt 3}\)

\(\vec a=\frac {λ}{3}(\hat i+\hat j+\hat k),\  λ>0\)

\(\frac {λ}{\sqrt 3}\frac {(\hat i+\hat j+\hat k)⋅(3\hat i+4\hat j)}{\sqrt {3^2+4^2}}=7\)

\(⇒\frac {λ}{\sqrt 3}(3+4)=7×5\)

∴ \(λ=5\sqrt 3\)

\(\vec a=5(\hat i+\hat j+\hat k)\)
Let
\(\vec b=p\hat i+q\hat j+r\hat k\)
\(\vec a⋅\vec b=0\) and \([\vec a \vec b \hat i]=0\)
\(⇒ p + q + r = 0\)       …(i)
And
\(\begin{vmatrix} p & q & r \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix}=0\)
\(⇒\vec b=−2r\hat i+r\hat j+r\hat k\)
\(⇒\vec b=r(−2\hat i+\hat j+\hat k)\)
Now
\(|\vec a|=|\vec b|\)
\(5\sqrt 3=|r|\sqrt b\)
\(|r|=5\sqrt 2\)
Projection of \(\vec b\) on \(3\hat i+4\hat j\)
= \(|\frac {\vec b⋅(3\hat i+4\hat j)}{\sqrt {3^2+4^2}}|\)

=\(|r|\frac {(−6+4)}{5}\)

=\(|−\frac 25|\)
Projection =\(\frac 25×\frac {5}{\sqrt 2}\)
Projection = \(\sqrt 2\)

So, the correct option is (b): \(\sqrt 2\)