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
distance of a point A() to line = + t is 
which is also equal to 
. Ans:
Let P is point on line such that
& P.V. of P is = + t . = 0 (
- ).( - ) = 0 ........................... (i)(
+ t - ).( + t - ) = 0 (
- + t).(t) = 0 t. - t + t2 = 0 t( - ) + t2c2 = 0 t = 
..................................... (ii) From (i) & (ii)
....................................... (iii) Also, PA2 = AB2 - BP2 = |
|2 - ||2
....................................... (iv) 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 +
l, m + m, n + n show that small anhgle 5
b/w two position is 2 = l2 + m2 + n2Ans: As we know l2 + m2 + n2 = 1
and (l +
l)2 + (m + m)2 + (n + n)2 = 1 l2 + (l2) 2ll + m2 + (m)2 + 2mm + n2 + (n)2 + 2nn = 1 l2 + m2 + n2 + (l)2 + (m)2 + (n)2 + 2(ll + mm + nn) = 1 (l)2 + (m)2 + (n)2 = - 2(ll + mm + nn) ............................ (i) is angle b/w two positions. cos
= l(l + l) + m(m + m) + n(n + n) 1 - 2 sin2
= 1 + ll + mm + nn ........................................... (ii) From (i) & (ii)
(
l)2 + (m)2 + (n)2 = 4 sin2 4()2 = (l)2 + (m)2 + (n)2| ()2 = (l)2 + (m)2 + (n)2 |
Dumb Question: How sin2
is equal to ()2 ? Ans: Since
is very small so, is very small So, sin
as 0 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
be image of point P in plane (i) Eq of P R is,
= r R(2r + 3, r + 5, r + 6) [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)
2(2r + 3) +(r + 5) + (r + 6) = 0 6r + 17 = 0
r = - 
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)
+ (y - 5)
+ (z - 6)
= 0 x2 + y2 + z2 + 
= 0 x2 + y2 + z2 + 
= 0 3x2 + 3y2 + 3z2 + 16x - 13y - 19z - 79 = 0 Key words
Direction COsines. Direction Ratios. Reflection of a point. Skew Lines. Shortest distance. Intersection of lines. Normal. FV Angle bisector. Tangency of plane to sphere.