Test Index

CBSE Class 12 Math 2013 Solved Paper

© examsnet.com
Question : 27
Total: 29
Find the vector equation of the plane passing through three points with position vectors
^
i
+
^
j
−2
^
k
 , 2
^
i
−
^
j
+
^
k
 and
^
i
+2
^
j
+
^
k
. Also, find the coordinates of the point of intersection of this plane and the line
→
r
 = 3
^
i
−
^
j
−
^
k
 + λ (2
^
i
−2
^
j
+
^
k
)
Solution:  
Let the position vectors of the three points be,
→
a
 =
^
i
+
^
j
−2
^
k
 ,
→
b
 = 2
^
i
−
^
j
+
^
k
 and
→
c
 =
^
i
+2
^
j
+
^
k
.
So, the equation of the plane passing through the points
→
a
,
→
b
 and
→
c
 is
(
→
r
−
→
a
).[(
→
b
−
→
c
)×(
→
c
−
→
a
)] = 0
⇒ [
→
r
−
^
i
+j^→+2
^
k
] . [
^
i
−3
^
j
×
^
j
+3
^
k
] = 0
⇒ [
→
r
−
^
i
+j^→+2
^
k
] .
^
k
−3
^
j
−9
^
i
 = 0
⇒
→
r
.(9
^
i
+3
^
j
−
^
k
) = 14 ... (1)
So, the vector equation of the required plane is
→
r
.(9
^
i
+3
^
j
−
^
k
) = 14
The equation of the given line is
→
r
 = 3
^
i
−
^
j
−
^
k
+λ(2
^
i
−2
^
j
+
^
k
)
Position vector of any point on the given line is
→
r
 = 3 + 2λ
^
i
 + (- 1 - 2λ)
^
j
 + (- 1 + λ)
^
k
 ... (2)
The point (2) lies on plane (1) if,
|(3+2λ)
^
i
+(−1−2λ)
^
j
+(−1+λ)
^
k
|.(9
^
i
+3
^
j
−
^
k
)
= 14
⇒ 9 (3) + 2λ + 3 - 1 - 2λ - (-1) + λ = 14
⇒ 11λ + 25 = 14
⇒ λ = - 1
Putting λ = - 1 in (2), we have
→
r
 = (3 + 2λ)
^
i
 + (- 1 - 2λ)
^
j
 + (- 1 + λ)
^
k
= (3 + 2 - 1)
^
i
 + (- 1 - 2 - 1)
^
j
 + (- 1 + (- 1))
^
k
=
^
i
+
^
j
−2
^
k
Thus, the position vector of the point of intersection of the given line and plane (1) is
^
i
+
^
j
−2
^
k
 and its co-ordinates are 1, 1, - 2 .
© examsnet.com
Go to Question:
Prev Question
Next Question