Question

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

Answer 1

($2\hat i+4\hat j-5\hat k$)+($\lambda \hat i+2\hat j+3\hat k$)
=(2+λ)$\hat i$+6$\hat j$-2$\hat k$
Therefore,unit vector along ($2\hat i+4\hat j-5\hat k$)+($\lambda \hat i+2\hat j+3\hat k$)is given as:
Scalar product of (i^+j^+k^)with this unit vector is 1.
$\Rightarrow$ ($\hat i+\hat j+\hat k$).(2+λ)$\hat i$+6$\hat j$-2$\hat k$/$\sqrt{\lambda^2+4\lambda+44}$=1
⇒$\frac{(2+\lambda)+6-2}{\lambda^2+4\lambda+44}$=1
⇒$\sqrt{\lambda^2+4\lambda+44}=\lambda+6$
⇒$\lambda^2+4\lambda+44=(\lambda+6)^2$
⇒$\lambda^2+4\lambda+44=\lambda^2+12\lambda+36$
⇒$8\lambda=8$
⇒λ=1
Hence, the value of λ is1.