3D Geometry - 7
On solving, we get 1 = 0 & µ1 = µ2
=
+
1(
-
) + µ1(
+
)
Since 1 = 0
=
+ µ1(
+
) is required eq. of straight line.
Q.3. Prove that three lines from O with direction cosines l1, m1, n1; l2, m2, n2; l3, m3, n3 are coplaner if
l1(m2n3 - n2m3) + m1(n2l3 - l2n3) + n1(l2m3 - l3m2) = 0
Ans: Note: Three given lines are coplanar if they have common perpendicular. Let d.c’s of common be l, m, n
ll1 + mm1 + nn1 = 0 ……………………….. (i)
ll2 + mm2 + nn2 = 0 ……………………….. (ii)
ll3 + mm3 + nn3 = 0 ……………………….. (iii)
Soplving (ii) & (iii) by cross multiplication ……
l = k(m2n3 - n2m3), m = k(n2l3 - n3l2), n = k(l2m3 - l3m2)
Substituting in (i), we get
k(m2n3 - n2m3)l1 + k(n2l3 - n3l2)m1 + k(l2m3 - l3m2)n1 = 0
l1(m2n3 - n2m3) + m1(n2l3 - n3l2) + n1(l2m3 - l3m2) = = 0
Q.4. Solve the equation x + y
+ z
=
Ans: x + y
+ z
=
……………………………. (i)
Taking dot product by x
, we get
x.(
x
) + y
.(
x
) + z
.(
x
) =
.(
x
)
x[] = [
]
x =
& z =
x
+ y
+ z
=
[
]
+ [
]
+ [
]
= [
]
is required solution.
Q.5. If planes x - cy - bz = 0, cx - y + az = 0 & bx + ay - z = 0 pass through a straight line, the find value of a2 + b2 + c2 + 2abc.
Ans: Givem planes are:
x - cy - bz = 0 ……………………….. (i)
cx - y + az = 0 ……………………….. (ii)
& bx + ay - z = 0 ……………………….. (iii)
Eq. of plane passing through line of intersection of plane (i) & (ii) is
(x - cy - bz) + (cx - y + az) = 0
x(1 +
c) -
(c +
) + z(- b + a
) = 0 ……………………….. (iv)
If plane (iv) & (iii) are same, then,
and
a - a3 + bc - a2bc = a2bc + ac2 + ab2 + bc
a(2abc + c2 + b2 + a2 - 1) = 0
a2 + b2 + c2 + 2abc = 1
Q.1. Show that







Ans:

Let P is point on line such that









(




(






(

















From (i) & (ii)



Also, PA2 = AB2 - BP2 = |





Dumb Question: How this eq. (iv) const.
Ans: We know that (AB)2 -

Q.2. If direction cosines of variable line in two adjacent points be l, m, n & l +









Ans: As we know l2 + m2 + n2 = 1
and (l +


























cos





1 - 2 sin2




From (i) & (ii)
(









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Dumb Question: How sin2


Ans: Since



So, sin






So, for approsmation we have taken.
Q.3. Find image of point P(3, 5, 6) in plane 2x + y + z = 0 & find eq. of sphere having point P(3, 5, 6) & its image point of extremities of diameter.
Ans:

2x + y + z = 0
P(3, 5, 6)
dr’s iof normal to plane (i) are 2, 1, 1
Let

Eq of P R is,


[Dumb Question: How eq. of line PR come ?
Ans: d.r’s of normal to plane is same as d.r’s of line PR i.e. 2, 1, 1
So, we get eq. of line PR]
But R lies on (i)

6r + 17 = 0


R

Since R lies mid point of PQ



Q =

Eq. of sphere when extremities (x1, y1, z1) & (x2, y2, z2) are given:
(x - x1)(x - x2) + (y - y1)(y - y2) + (z - z1)(z - z2) = 0
(x - 3)








Key words








