Chapter 6 Line and Plane Miscellaneous Exercise 6B Solutions

(b)

(a)

(A) 4, 5, 7

(C )

(c)

(B ) k = 0 or -3

(B) inrersecting

(C) y = 0, z = 0

(D ) 90º

(B ) 2, 1, -6

(A ) 14

(d) cos-1

(d) , 2

(D ) x + y + z = 4

(A ) 0º

(a)

(B ) 3x – 2z = 1

(A ) 17x – 47y – 24z + 172 = 0

(A ) 5

(B ) 4x – 2y – 5z = 45

The distance of the point (x1, y1, z1) from the plane ax + by + cz + d is ∴ the distance of the point (1, 1, -1) from the plane 6x + 2y + 3z – 7 = 0 is = 1units.

The equation of the plane is 2x + 3y + 6z = 49 Dividing each term by = = 7 we get x + y – z = = 7 This is the normal form of the equation of plane. ∴ the direction cosines of the perpendicular drawn from the origin to the plane are l = , m = , n = and length of perpendicular from origin to the plane is p = 7. the coordinates of the foot of the perpendicular from the origin to the plane are (lp, ∓, np)i.e.(2, 3, 6)

The normal form of equation of a plane is = p where is unit vector along the normal and p is the length of perpendicular drawn from origin to the plane. Given pane is …(1) This is the normal form of the equation of plane. Comparing with = p, (i) the length of the perpendicular from the origin to plane is . (ii) direction cosines of the normal are

The vector equation of the plane passing through three non-collinear points A(), B() and C() is … (1)

The Cartesian equation of the plane passing through (x1, y1, z1), the direction ratios of whose normal are a, b, c, is a(x – x1) + b(y – y1) + c(z – z1) = 0 ∴ the cartesian equation of the required plane is o(x + 1) + 2(y + 2) + 5(z – 3) = 0 i.e. 0 + 2y – 4 + 10z – 15 = 0 i.e. y + 2 = 0.

The cartesian equation of the plane is 6x + 8y + 7z = 0 The required plane is parallel to it ∴ its cartesian equation is 6x + 8y + 7z = p …(1) A (7, 8, 6) lies on it and hence satisfies its equation ∴ (6)(7) + (8)(8) + (7)(6) = p i.e., p = 42 + 64 + 42 = 148. ∴ from (1), the cartesian equation of the required plane is 6x + 8y + 7z = 148.

The vector equation of the plane passing through A() and perpendicular to is . M(1, 2, 0) is the foot of the perpendicular drawn from origin to the plane. Then the plane is passing through M and is perpendicular to OM. If is the position vector of M, then = . Normal to the plane is = = = = 5 ∴ the vector equation of the required plane is = 5

The vector equation of the plane passing through A(), B().. C(), where A, B, C are non collinear is …(1) The required plane makes intercepts 1, 1, 1 on the coordinate axes. ∴ it passes through the three non collinear points A = (1, 0, 0), B = (0, 1, 0), C = (0,, 1)

The vector equation of the plane passing through the point A() and parallel to the vectors and is = (-2)(-4) + (7)(-1) + (5)(4) = 8 – 7 + 8 = 35 ∴ From (1), the vector equation of the required plane is = 35.

The equation represents a plane passing through a point having position vector and parallel to vectors and . Here,

Case 1 : Let all the intercepts be 0. Then the plane passes through the origin. Then the cartesian equation of the plane is ax + by + cz = 0 …..(1) (1, 2, 3) and (3, 2, 1) lie on the plane. ∴ a + 2b + 3c = 0 and 3a + 2b + c = 0 ∴ a, b, c are proportional to 1, -2, 1 ∴ from (1), the required cartesian equation is x – 2y + z = 0. Case 2 : Let the plane make non zero intercept p on each axis. then its equation is = 1 i.e. x + y + z = p …(2) Since this plane pass through (1, 2, 3) and (3, 2, 1) ∴ 1 + 2 + 3 = p and 3 + 2 + 1 = p ∴ p = 6 ∴ from (2), the required cartesian equation is x + y + z = 6 Hence, the cartesian equations of required planes are x + y + z = 6 and x – 2y + z = 0.

Case 1 : Let all the intercepts be 0. Then the plane passes through the origin. Then the vector equation of the plane is ax + by + cz …(1) (1, 1, 1) lie on the plane. ∴ 1a + 1b + 1c = 0 ∴ from (1), the required cartesian equation is x – y + z = 0 Case 2 : Let he plane make non zero intercept p on each axis. then its equation is = 1 i.e. = p ….(2) Since this plane pass through (1, 1, 1) ∴ 1 + 1 + 1 = p ∴ p = 3 ∴ from (2), the required cartesian equation is = 3 Hence, the cartesian equations of required planes are

The acute angle between the planes = (1)(2) + (1)(1) + (2)(1) = 2 + 1 + 2 = 5 Also,

The acute angle θ between the line and the plane = d is given by = (2)(2) + (3)(-1) + (-6)(1) = 4 – 3 – 6 = -5

The distance of the point A() from the plane = p is given by d = ……(1) = (3)(2) + (3)(3) + (1)(-6) = 6 + 9 – 6 = 9

= 19units.

The vector equation of the plane passing through A() and perpendicular to the vector is … (1) We can take = since the plane passes through the origin. The point M with position vector = lies on the line and hence it lies on the plane..’. lies on the plane. The plane contains the given line which is parallel to Let be normal to the plane. Then is perpendicular to as well as

The vector equation of the plane passing through A() and perpendicular to the vector is ….(1) The position vectors and of the given points A and B are and If M is the midpoint of segment AB, the position vector of M is given by

Given lines are x = y, z = 0 and x + y = 0, z = 0. It is clear that (0, 0, 0) satisfies both the equations. ∴ the lines intersect at O whose position vector is Since z = 0 for both the lines, both the lines lie in XY- plane. Hence, we have to find equation of XY-plane. Z-axis is perpendicular to XY-plane. ∴ normal to XY plane is . 0() lies on the plane. By using , the vector equation of the required plane is i.e. . Hence, the given lines intersect each other and the vector equation of the plane determine by them is .