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Q1. Find the angle between any two diagonals of a cube.

Solution

 Let O, one vertex of a cube, be the origin and three edges through O be the Co-ordinate axes. The four diagonals are OP, AA', BB' and CC'. Let 'a' be the length of each edge. Then the co-ordinates of P, A, A' are (a, a, a), (a, 0, 0), (0, a, a).             The direction ratios of OP are a, a, a.             The direction cosines of OP are  .             Similarly, direction cosines of AA' are  .             Let be the angle between the diagonals OP and AA'. Then                                          . Thus the angle between any two diagonals of a cube is .  

Q2. Find the equation of the plane through the intersection of the plane , and perpendicular to the plane .

Solution

The equation of the required plane is:  

Q3. Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 6z + 7 = 0.

Solution

Let the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 6z + 7 = 0 is (x₁, y₁, z₁). Now, the equation in the normal form is: Substituting these in the equation of the plane, we get Hence, the foot of the perpendicular is    

Q4. Find the equation of the plane passing through the line of intersection of the planes 3x – 5y + 4z + 11 = 0, 2x – 7y + 4z – 3 = 0 and the point (– 2, 1, 3).

Solution

The equation of the plane passing through the line of intersection of the planes 3x – 5y + 4z + 11 = 0, 2x – 7y + 4z – 3 = 0 is: . Also, the plane passing through the point (– 2, 1, 3), Now, the required equation of the plane is:

Q5. Find the equation of the line in vector and in Cartesian form that passes through the point with position vector  and is in the direction .

Solution

The vector form of the equation passes through  and parallel to the vectoris: . The Cartesian form of this equation is: .    

Q6. Find the equation of the plane passing through the points (2, 5, – 8) and perpendicular to each one of the planes 2x – 3y + 4z + 1 = 0 and 4x + y – 2z + 6 = 0.

Solution

Equation of the plane passing through the point (2, 5, – 8) is: a(x – 2) + b(y – 5) + c(z + 8) = 0                                                        …(1) If the plane perpendicular to the plane 2x – 3y + 4z + 1 = 0, then 2a – 3b + 4c = 0                                                                                  …(2) and If the plane perpendicular to the plane 4x + y – 2z + 6 = 0, then 4a + b – 2c = 0                                                                                    …(3) On solving equations (2) and (3), we get On substituting the proportional values of a, b and c in (1), we get (x – 2) + 10(y – 5) + 7(z + 8) = 0

Q7. Find the direction cosines of the line segment joining the points (2, 0, 1) and (–1, 3, –2).

Solution

The direction ratios of the line segment joining the points (2, 0, 1) and (–1, 3, –2) are (– 1 – 2), (3 – 0), (– 2 – 1) i.e., –3, 3, –3.    

Q8. Find the equation of the plane passing through the points (–1, 4, – 3), (3, 2, – 5) and (– 3, 8, – 5).

Solution

 Let . The equation of the plane is:  

Q9. Show that the points A (2, 3, - 4), B (1, - 2, 3) and C (3, 8, - 11) are collinear.

Solution

Direction ratios of line joining A and B are 1 – 2, – 2 – 3, 3 + 4 i.e., – 1, – 5, 7. The direction ratios of line joining B and C are 3 –1, 8 + 2, – 11 – 3, i.e., 2, 10, – 14. It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel to BC. But point B is common to both AB and BC. Therefore, A, B, C are collinear points.    

Q10. Find the shortest distance between the lines: and .

Solution

The given equations can be written as

Q11. If a line makes angles  with OX, OY and OZ respectively. Prove that .

Solution

Let be the direction cosines of the given line. Then .

Q12. Find the Cartesian equation of line which passes through the points (2, 0, 5) and (4, – 3, – 2).

Solution

 Cartesian equation of line passing through the points (2, 0, 5) and (4, – 3, – 2) is:

Q13.

Solution

Q14. The foot of the perpendicular drawn from the origin to a plane is (2, 1, 5). Find the equation of the plane.

Solution

Since, the foot of the perpendicular to the plane is A(2, 1, 5). Therefore, (2, 1, 5) is the point on the plane. So, equation of the plane passing through the point (2,.1, 5) is: a(x – 2) + b(y – 1) + c(z – 5) = 0. Now, the direction ratios of the perpendicular line OA = 2 – 0, 1 – 0, 5 – 0, i.e., 2, 1, 5.   Therefore, the required plane is: 2(x – 2) + 1(y – 1) + 5(z – 5) = 0 i.e, 2x + y + 5z = 30.    

Q15. Find the equation of the plane which passes through the point (3,4,-1) and is parallel to the plane 2x - 3y + 5z + 7 = 0. Also find the distance between the two planes.

Solution

Q16. Show that the lines and are coplanar.

Solution

Here, x₁ = 1, y₁ = – 2, z₁ = – 5, a₁ = 1, b₁ = – 2, c₁ = – 5 and x₂ = 3, y₂ = – 1, z₂ = – 5, a₂ = 3, b₂ = – 1, c₂ = – 5. Therefore, the lines are coplanar.  

Q17. If a line makes angles  with OX, OY and OZ respectively. Prove that .

Solution

 Let be the direction cosines of the given line. Then .    

Q18. Find the Cartesian equation of the line which passes through the origin and parallel to the line .

Solution

The direction ratios of the line, which parallel to the lineis (– 1, – 2, 5). The Cartesian equation of the line which passes through the origin and parallel to the line is .  

Q19. Find the shortest distance between the lines:  and .

Solution

Q20. The vertices of a triangle ABC are A (–1, 2, –3), B (5, 0, –6) and C (0, 4, –1). Find the direction cosines of the bisector of the angle BAC.

Solution

 Here, AB =. AC = .   By geometry, the bisector of will divide the side BC in the ratio AB : AC i.e., in the ratio 7 : 3 internally. Let the bisector of meets the side BC at point D.   Therefore, D divides BC in the ratio 7 : 3.        Coordinates of D are   i.e., .        Therefore, direction ratios of the bisector AD are.        Hence, direction cosines of the bisector AD are , , i.e.,. 

Q21. Find the angle between the line  and the plane 5x – 4y + 7z + 10 = 0.

Solution

 Let  be the angle between the given line and the plane. The vector equations of the line and plane are: and. Here, and. Now, angle between the line and the plane is given by    

Q22. A variable plane is at a constant distance p from the origin and meets the axes in points A,B,C. Through A,B, C planes are drawn parallel to the coordinate planes. Prove that the locus of their point of intersection is x^-2 + y^-2+z^-2 = p^-2.

Solution

Q23. Show that the lines and are perpendicular.

Solution

The direction ratios of the first line are 3, – 2, 5 and the direction ratios of the second line are 1, – 1, -1. If the angle between the lines is, then   Thus, the lines are perpendicular.

Q24. Find the distance of the point (3, – 2, 5) from the plane .

Solution

Here, ,  and d = 5. Now, the distance of the point (3, – 2, 5) from the plane is: