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Practice each question, then open the answer or worked solution.
Find the position vector of a point A in space such that 
is inclined at 60º to OX and at 45° to OY and 
= 10 units.
We know that, cos2 α + cos2 β + cos2 γ = 1
cos2 60o + cos2 45o + cos2 γ = 1
(1/2)2 + (1/√2)2 + cos2 γ = 1
¼ + ½ + cos2 γ = 1
cos2 γ = 1 – ¾ = ¼
So, cos γ = ± ½ ⇒ cos γ = ½ [Rejecting cos γ = – ½, as γ < 90o]
Now,
Find the vector equation of the line which is parallel to the vector 
and which passes through the point (1, –2, 3).
We know that the equation of line is
Show that the lines
Also, find their point of intersection.
Given equation are,
Find the angle between the lines
Prove that the line through A (0, –1, –1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (– 4, 4, 4).
Given points, A (0, –1, –1) and B (4, 5, 1)
C (3, 9, 4) and D (– 4, 4, 4).
Cartesian form of equation AB is
Prove that the lines x = py + q, z = ry + s and x = p¢y + q¢, z = r¢y + s¢ are perpendicular if pp¢ + rr¢ + 1 = 0.
Find the equation of a plane which bisects perpendicularly the line joining the points A (2, 3, 4) and B (4, 5, 8) at right angles.
Given coordinates are A (2, 3, 4) and B (4, 5, 8)
Now, the coordinates of the mid-point C are (2+4/2, 3+5/2, 4+8/2) = (3, 4, 6)
And, the direction ratios of the normal to the plane = direction ratios of AB
= 4 – 2, 5 – 3, 8 – 4 = (2, 2, 4)
Equation of the plane is
a(x – x1) + b(y – y1) + c(z – z1) = 0
2(x – 3) + 2(y – 4) + 4(z – 6) = 0
2x – 6 + 2y – 8 + 4z – 24 = 0
2x + 2y + 4z = 38
x + y + 2z = 19
Thus, the required equation of plane is x + y + 2z = 19 or
Find the equation of a plane which is at a distance 3√3 units from origin and the normal to which is equally inclined to coordinate axis.
As the normal to the plane is equally inclined to the axes we have,
cos α = cos β = cos γ
So, cos2 α + cos2 α + cos2 α = 1
3 cos2 α = 1 ⇒ cos α = ± 1/√3
And, cos α = cos β = cos γ = ± 1/√3
Thus, the equation of the plane is x + y + z = 9
If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane.
Given, points (–2, – 1, – 3) and (1, – 3, 3)
Direction ratios of the normal to the plane are (1 + 2, -3 + 1, 3 + 3) = (3, -2, 6)
Now, the equation of plane passing through one point (x1, y1, z1) is
a(x – x1) + b(y – y1) + c(z – z1) = 0
3(x – 1) – 2(y + 3) + 6(z – 3) = 0
3x – 3 – 2y – 6 + 6z – 18 = 0
3x – 2y + 6z – 27 = 0 ⇒ 3x – 2y + 6z = 27
Thus, the required equation of plane is 3x – 2y + 6z = 27.
Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).
Given points are (2, 1, 0), (3, –2, –2) and (3, 1, 7)
As the equation of the plane passing through the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is
(x – 2) (-21) – (y – 1)(7 + 2) + z (3) = 0
-21 (x – 2) – 9(y – 1) + 3z = 0
-21x + 42 – 9y + 9 + 3z = 0
-21x – 9y + 3z + 51 = 0 ⇒ 7x + 3y – z – 17 = 0
Thus, the required equation of plane is 7x + 3y – z – 17 = 0.