Question

The scalar product of the vector $i+j+\hat{k}$ with a unit vector along the sum of vectors $2i+4j-5\hat{k}$ and
$\lambda i+2j+3\tilde{k}$ is equal to one. The value of $\lambda$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Let,

$a=i+j+k$

$b=2i+4j-5k$

$c=\lambda i+2j+3k$

$b+c=2i+4j-5k+\lambda i+2j+3k=(2+\lambda )i+6j-2k$

Let "n" be the unit vector along (b+c)

$n=\frac{1}{|b+c|}\times b+c$

$n=\frac{1}{\sqrt{(2+\lambda {)}^2+6^2+(-2{)}^2}}\times (2+\lambda )i+6j-2k$

$\therefore n=\frac{1}{\sqrt{{\lambda }^2+4\lambda +44}}\times (2+\lambda )i+6j-2k$

Given $a\cdot n=1$

we have,

$(i+j+k)\cdot (\frac{1}{\sqrt{{\lambda }^2+4\lambda +44}}\times (2+\lambda )i+6j-2k)=1$

$(i+j+k)\cdot \left(\right(2+\lambda )i+6j-2k)=\sqrt{{\lambda }^2+4\lambda +44}$

$\lambda +2+6-2=\sqrt{{\lambda }^2+4\lambda +44}$

$\lambda +6=\sqrt{{\lambda }^2+4\lambda +44}$

squaring on both sides, we have,

$(\lambda +6{)}^2={\lambda }^2+4\lambda +44$

${\lambda }^2+36+12\lambda ={\lambda }^2+4\lambda +44$

$8\lambda =8$

$\therefore \lambda =1$