The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors…
The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to 1 , then value of $\lambda$ is
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Solution
$(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})+(\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=(2+\lambda) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
Unit vector along the above vector is
$\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}}$
Scalar product of $(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ with this unit vector is 1 .
$\begin{aligned}
\therefore \quad & (\hat{i}+\hat{j}+\hat{k}) \cdot \frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}}=1 \\
& \frac{(2+\lambda)+6-2}{\sqrt{\lambda^2+4 \lambda+44}}=1 \\
& \sqrt{\lambda^2+4 \lambda+44}=\lambda+6 \\
\therefore \quad & \lambda^2+4 \lambda+44=(\lambda+6)^2 \\
& 8 \lambda=8 \\
& \lambda=1
\end{aligned}$