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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 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 .

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Let, the sum of vectors$2\hat i + 4 \hat j -5 \hat k$  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,

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

$\lambda =1$

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