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CBSE Class 12 Math 2018 Solved Paper

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Question : 28 of 29
Marks: +1, -0
Find the distance of the point (-1,-5,-10) from the point of intersection of the line
r⃗\vec{r} = 2i^−j^+2k^\hat{2i} - \hat{j} + \hat{2k} + λ (3i^+4j^+2k^)(\hat{3i} + \hat{4j} + \hat{2k}) and the plane r⃗\vec{r} = (i^−j^+k^)(\hat{i} - \hat{j} + \hat{k}) = 5
Solution:  
Equation of line is r⃗\vec{r} = 2i^−j^+2k^\hat{2i} - \hat{j} + \hat{2k} + λ (3i^+4j^+2k^)(\hat{3i} + \hat{4j} + \hat{2k})
r⃗\vec{r} = (2 + 3λ) i^\hat{i} + (- 1 + 4λ) j^\hat{j} + (2 + 2λ) k^\hat{k}
This lies on the plane r⃗\vec{r} = (i^−j^+k^)(\hat{i} - \hat{j} + \hat{k}) = 5
⇒ [(2 + 3λ) i^\hat{i} + (- 1 + 4λ) j^\hat{j} + (2 + 2λ) k^\hat{k}] . (i^−j^+k^)(\hat{i} - \hat{j} + \hat{k}) = 5
⇒ (2 + 3λ) + (- 1 + 4λ) (- 1) + (2 + 2λ) = 5
⇒ λ = 0
Coordinates are
(2 + 3λ , - 1 + 4λ , 2 + 2λ) = (2 , - 12)
Distance between (2 , - 1 , 2) and (- 1 , - 5 , - 10)
=
(−1−2)2+(−5−1)2+(−10−2)2\sqrt{(-1-2)^2+(-5-1)^2+(-10-2)^2}
= 13 units
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