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  • The scalar product of the vector iˆ + ˆj + kˆ with a unit vector along the sum of vectors 2iˆ + 4 ˆj − 5kˆ and iˆ + 2 ˆj + 3kˆ is equal to one. Find the value of .

Q13  The scalar product of the vector \hat i + \hat j + \hat k with a unit vector along the sum of vectors2\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 .

Answers (1)

best_answer

Let, the sum of vectors2\hat i + 4 \hat j -5 \hat k  and \lambda \hat i + 2 \hat j +3 \hat k be 

\vec a=(\lambda +2)\hat i + 6 \hat j -2 \hat k

unit vector along \vec a

\vec u=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{(\lambda+2)^2+6^2+(-2)^2}}=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}

Now, the scalar product of this with \hat i + \hat j + \hat k

\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2)\hat i + 6 \hat j -2 \hat k}{\sqrt{\lambda^2+4\lambda+44}}.(\hat i+\hat j +\hat k)

\vec u.(\hat i+\hat j +\hat k)=\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1

\frac{(\lambda +2) + 6 -2 }{\sqrt{\lambda^2+4\lambda+44}}=1

\frac{(\lambda +6) }{\sqrt{\lambda^2+4\lambda+44}}=1

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

squaring both the side,

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

\lambda =1

Posted by

Pankaj Sanodiya

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