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.