Test Index

CBSE Class 12 Math 2018 Solved Paper

© examsnet.com
Question : 28
Total: 29
Find the distance of the point (-1,-5,-10) from the point of intersection of the line
→
r
 = 2
^
i
−
^
j
+2
^
k
 + λ (3
^
i
+4
^
j
+2
^
k
) and the plane
→
r
 = (
^
i
−
^
j
+
^
k
) = 5
Solution:  
Equation of line is
→
r
 = 2
^
i
−
^
j
+2
^
k
 + λ (3
^
i
+4
^
j
+2
^
k
)
→
r
 = (2 + 3λ)
^
i
 + (- 1 + 4λ)
^
j
 + (2 + 2λ)
^
k
This lies on the plane
→
r
 = (
^
i
−
^
j
+
^
k
) = 5
⇒ [(2 + 3λ)
^
i
 + (- 1 + 4λ)
^
j
 + (2 + 2λ)
^
k
] . (
^
i
−
^
j
+
^
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

= 13 units
© examsnet.com
Go to Question:
Prev Question
Next Question