**Chapter 5 Vectors Ex 5.1**

## Chapter 5 Vectors Ex 5.1

**Question 1.**

Solution:

Solution:

**Question 3.Solution:**

**Question 4.If ABCDEF is a regular hexagon, show Solution:**

ABCDEF is a regular hexagon.

∴ by the triangle law of addition of vectors,

**Question 5.**

**Question 6.Solution:**

**Question 7.Find the distance from (4, -2, 6) to each of the following :(a) The XY-planeSolution:**

Let the point A be (4, -2, 6).

Then,

The distance of A from XY-plane = |z| = 6

**(b) The YZ-planeSolution:**

The distance of A from YZ-plane = |x| = 4

**(c) The XZ-planeSolution:**

The distance of A from ZX-plane = |y| = 2

**(d) The X-axisSolution:**

The distance of A from X-axis

**(e) The Y-axisSolution:**

The distance of A from Y-axis

**(f) The Z-axisSolution:**

The distance of A from Z-axis

**Question 8.Find the coordinates of the point which is located :(a) Three units behind the YZ-plane, four units to the right of the XZ-plane and five units above the XY-plane.Solution:**

Let the coordinates of the point be (x, y, z).

Since the point is located 3 units behind the YZ- j plane, 4 units to the right of XZ-plane and 5 units , above the XY-plane,

x = -3, y = 4 and z = 5

Hence, coordinates of the required point are (-3, 4, 5)

**(b) In the YZ-plane, one unit to the right of the XZ-plane and six units above the XY-plane.Solution:**

Let the coordinates of the point be (x, y, z).

Since the point is located in the YZ plane, x = 0. Also, the point is one unit to the right of XZ-plane and six units above the XY-plane.

∴ y = 1, z = 6.

Hence, coordinates of the required point are (0, 1, 6).

**Question 9.Find the area of the triangle with vertices (1, 1, 0), (1, 0, 1) and (0, 1, 1).Solution:**

Let A = (1, 1, 0), B = (1, 0, 1), C = (0, 1, 1)

**Question 10.**

**Question 11.Show that the following points are collinear :(i) A (3, 2, -4), B (9, 8, -10), C (-2, -3, 1).Solution:**

A = (3, 2, -4), B = (9, 8, -10) and C = (-2, -3, 1) respectively.

**(ii) P (4, 5, 2), Q (3, 2, 4), R (5, 8, 0).Solution:**

P = (4, 5, 2), Q = (3, 2, 4), R = (5, 8, 0) respectively.

**Question 12.**

**Question 13.Are the four points A(1, -1, 1), B(-1, 1, 1), C(1, 1, 1) and D(2, -3, 4) coplanar? Justify your answer.Solution:**

By equality of vectors,

y = -2 ….(1)

2x – 2y = 2 … (2)

3y = 0 … (3)

From (1), y = -2

From (3), y = 0 This is not possible.

Hence, the points A, B, C, D are not coplanar.

**Question 14.**

**Solution:**

By equality of vectors,

2x + 2y + 3 = -1

x – y + z = -3

-4x + 3y – 2z = 4

We have to solve these equations by using Cramer’s Rule

= 2(2 – 3) – 2(-2 + 4) + 3(3 – 4)

= -2 – 4 – 3 = -9 ≠ 0

= 2(-4 + 9) – 2(4 – 12) – 1(3 – 4)

= 10 + 16 + 1 = 27