**Chapter 5 Vectors Ex 5.3**

## Chapter 5 Vectors Ex 5.3

**Question 1.**

**Question 2.**

âˆ´ LHS = RHS

Hence, (aÂ¯ + bÂ¯)

^{2}= (aÂ¯ â€“ bÂ¯)

^{2}.

**Question 3.**

âˆ´ c < 0.

Hence, the angle between a and b is obtuse if c < 0.

**Question 4.**

**Question 5.Suppose that all sides of a quadrilateral are equal in length and opposite sides are parallel. Use vector methods to show that the diagonals are perpendicular.Solution:**

**Question 6.**

= -3 + 2 + 1

= 0

**Question 7.Find the angle P of the triangle whose vertices are P(0, -1, -2), Q(3, 1, 4) and R(5, 7, 1).Solution:**

âˆ´ P = 45Â°

**Question 8.**

Solution:

**Question 9.Prove by vector method that the angle subtended on semicircle is a right angle.Solution:**

Hence, the angle subtended on a semicircle is the right angle.

**Question 10.If a vector has direction angles 45Âºand 60Âº find the third direction angle.Solution:**

Let Î± = 45Â°, Î² = 60Â°

We have to find Î³.

âˆ´ cos

^{2}Î± + cos

^{2}Î² + cos

^{2}Î³ = 1

âˆ´ cos

^{2}45Â° + cos

^{2}60Â° + cos

^{2}Î³ = 1

**Question 11.If a line makes angles 90Âº, 135Âº, 45Âº with the X, Y and Z axes respectively, then find its direction cosines.Solution:**

Let l, m, n be the direction cosines of the line.

Then l = cos Î±, m = cos Î², n = cos Î³

Here, Î± = 90Â°, Î² = 135Â° and Î³ = 45Â°

âˆ´ l = cos 90Â° = 0

**Question 12.If a line has the direction ratios, 4, -12, 18 then find its direction cosines.Solution:**

The direction ratios of the line are a = 4, b = -12, c = 18.

Let l, m, n be the direction cosines of the line.

**Question 13.**

The coordinates of the points which are at a distance of d units from the point (x_{1}, y_{1}, z_{1}) are given by (x_{1} Â± ld, y_{1} Â± md, z_{1} Â± nd)

i.e. (4 â€“ 4, 1 + 4, 5 + 2) and (4 + 4, 1 â€“ 4, 5 â€“ 2)

i.e. (0, 5, 7) and (8, -3, 3).

**Question 14.Find the angle between the lines whose direction cosines l, m, n satisfy the equations 5l + m + 3n = 0 and 5mn â€“ 2nl + 6lm = 0.Solution:**
Given, 5l + m + 3n = 0 â€¦(1)

and 5mn â€“ 2nl + 6lm = 0 â€¦(2)

From (1), m = -(51 + 3n)

Putting the value of m in equation (2), we get,

-5(5l + 3n)n â€“ 2nl â€“ 6l(5l + 3n) = 0

âˆ´ -25ln â€“ 15n

^{2}â€“ 2nl â€“ 30l

^{2}â€“ 18ln = 0

âˆ´ â€“ 30l

^{2}â€“ 45ln â€“ 15n

^{2}= 0

âˆ´ 2l

^{2}+ 3ln + n

^{2}= 0

âˆ´ 2l

^{2}+ 2ln + ln + n

^{2}= 0

âˆ´ 2l(l + n) + n(l + n) = 0

âˆ´ (l + n)(2l + n) = 0

âˆ´ l + n = 0 or 2l + n = 0

l = -n or n = -2l

Now, m = -(5l + 3n), therefore, if l = -n,

m = -(-5n + 3n) = 2n

âˆ´ the direction ratios of the second line are

a

_{2}= -1, b

_{2}= 1, c

_{2}= -2

Let Î¸ be the angle between the lines.