The distance of the point Q(0,2,-2) form the line passing through the point P(5,-4, 3) and…

Question

The distance of the point $Q(0,2,-2)$ form the line passing through the point $P(5,-4, 3)$ and perpendicular to the lines $$\vec{r}=(-3 \hat{i}+2 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+5 \hat{k}), \lambda \in \mathbb{R}$$

Accepted Answer

A vector in the direction of the required line can be obtained by cross product of

$$\begin{aligned}
& \left|\[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 5 \ -1 & 3 & 2 \end{array}\end{aligned}$$\right| \\
\& =-9 \hat{i}-9 \hat{j}+9 \hat{k}
\]

Required line

$$\begin{aligned}
& \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda^{\prime}(-9 \hat{\mathrm{i}}-9 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}) \\
& \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})
\end{aligned}$$

Now distance of $(0,2,-2)$

$$\begin{aligned}
& \text { P.V. of } \mathrm{P} \equiv(5+\lambda) \hat{i}+(\lambda-4) \hat{j}+(3-\lambda) \hat{k} \\
& \overrightarrow{\mathrm{AP}}=(5+\lambda) \hat{i}+(\lambda-6) \hat{j}+(5-\lambda) \hat{k} \\
& \overrightarrow{\mathrm{AP}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\
& 5+\lambda+\lambda-6-5+\lambda=0 \\
& \lambda=2 \\
& |\overrightarrow{\mathrm{AP}}|=\sqrt{49+16+9} \\
& |\overrightarrow{\mathrm{AP}}|=\sqrt{74}
\end{aligned}$$

Solutions · 1