Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line r→=2ı^−ȷ^+2k^+λ(3ı^+4ȷ^+2k^) and the plane r→⋅(ı^−ȷ^+k^)=5.

Given: r→=2ı^−ȷ^+2k^+λ(3ı^+4ȷ^+2k^)…… (i) Coordinates of any point on this line are (2+3λ)i^+(−1+4λ)j^+(2+2λ)k^Equation of plane is r→⋅(i^−j^+k^)=5 .....(ii)Since, the point on line lies on the plane, r→⋅(i^−i^+k^)=5[(2+3λ)i^+(2+4λ)j^+(2+2λ)](i^−j^+k^)=5⇒ (2+3λ)−(−1+4λ)+(2+2λ)=5⇒ λ+5=5⇒ λ=0 So equation of line isr→=(2i^−j^+2k^)+0(3i^+4j^+2k^)⇒r→=2i^−j^+2k^ ....(iii)Let, point of intersection be (x, y, z)∴ r→=xi^+yj^+zk .....(iv)^From equations (iii) and (iv)x=2,y=−1,z=2∴ Point of intersection is (2,-1,2).Distance between points (2,-1,2) and (-1,-5,-10) is =(−1−2)2+(−5+1)2+(−10−2)2=(−3)2+(−4)2+(−12)2=9+16+144=169=13 units Therefore, the distance of the point (– 1, – 5, – 10) from the point of intersection of the line r→=2ı^−ȷ^+2k^+λ(3ı^+4ȷ^+2k^) and the plane r→⋅(ı^−ȷ^+k^)=5 is 13 units.

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Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line $\vec{r} = 2 \hat{ı} - \hat{ȷ} + 2 \hat{k} + λ (3 \hat{ı} + 4 \hat{ȷ} + 2 \hat{k})$ and the plane $\vec{r} \cdot (\hat{ı} - \hat{ȷ} + \hat{k}) = 5$ .

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Detailed Solution

Given: $\vec{r} = 2 \hat{ı} - \hat{ȷ} + 2 \hat{k} + λ (3 \hat{ı} + 4 \hat{ȷ} + 2 \hat{k}) \dots \dots (i)$
Coordinates of any point on this line are
$(2 + 3 λ) \hat{i} + (- 1 + 4 λ) \hat{j} + (2 + 2 λ) \hat{k}$
Equation of plane is
$\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 . . . . . (ii)$
Since, the point on line lies on the plane,
$\vec{r} \cdot (\hat{i} - \hat{i} + \hat{k}) = 5$
$\begin{array}{l} [(2 + 3 λ) \hat{i} + (2 + 4 λ) \hat{j} + (2 + 2 λ)] (\hat{i} - \hat{j} + \hat{k}) = 5 \\ \Rightarrow (2 + 3 λ) - (- 1 + 4 λ) + (2 + 2 λ) = 5 \\ \Rightarrow λ + 5 = 5 \\ \Rightarrow λ = 0 \end{array}$
So equation of line is
$\begin{array}{l} \vec{r} = (2 \hat{i} - \hat{j} + 2 \hat{k}) + 0 (3 \hat{i} + 4 \hat{j} + 2 \hat{k}) \\ \Rightarrow \vec{r} = 2 \hat{i} - \hat{j} + 2 \hat{k} . . . . (iii) \end{array}$
Let, point of intersection be $(x, y, z)$
$∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k . . . . . (iv)}$
From equations (iii) and (iv)
$x = 2, y = - 1, z = 2$
$∴$ Point of intersection is (2,-1,2).
Distance between points (2,-1,2) and (-1,-5,-10) is
$\begin{array}{l} = \sqrt{(- 1 - 2)^{2} + (- 5 + 1)^{2} + (- 10 - 2)^{2}} \\ = \sqrt{(- 3)^{2} + (- 4)^{2} + (- 12)^{2}} \\ = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 units \end{array}$
Therefore, the distance of the point (– 1, – 5, – 10) from the point of intersection of the line
$\vec{r} = 2 \hat{ı} - \hat{ȷ} + 2 \hat{k} + λ (3 \hat{ı} + 4 \hat{ȷ} + 2 \hat{k})$ and the plane $\vec{r} \cdot (\hat{ı} - \hat{ȷ} + \hat{k}) = 5$ is 13 units.