Question
If the vectors $3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}$ are perpendicular, then $\lambda$ is equal to:
- $-14$
- $7$
- $14$
- $\frac{1}{7}$
Solution:
It is given that vectors $3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}$ are perpendicular.
So, their dot product is zero.
$\big(3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(2\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}\big)=0$
$\Rightarrow6-\lambda+8=0$
$\Rightarrow14-\lambda=0$
$\therefore\lambda=14$
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If $\mathrm{y}\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2},$ then $\mathrm{y}\left(\frac{-1}{\sqrt{2}}\right)$ is equal to
$\sqrt{1-\pi^2}-1$
$\frac{\pi}{2}-1$
$\frac{\pi}{2}+1$
${\pi}+{1}$
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