CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Vector Algebra

10.1. Types of Vector.

Q.1. If P→(1, 5, 4) and Q→(4, – 1, – 2), find the direction ratios of PQ→.

Solution :

We have, OP→ = i + 5j + 4k and OQ→ = 4i – j – 2k.
Therefore, PQ→ = OQ→ – OP→
= 4i – j – 2k – i – 5j – 4k
= 3i – 6j – 6k.
Therefore, Direction ratios = 3, – 6, – 6. [Ans.]

Q. 2. Find the magnitude of the vector : a→ = 2i – 6j – 3k.

Solution :

We have, a→ = 2i – 6j – 3k.
|a→| = √{(2)² + (– 6)² + (– 3)²}
= √(4 + 36 + 9)
= √(49)
= 7. [Ans.]

10. 2. Addition of Vectors.

Q.1. Using vectors, prove that the line segment joining the mid-points of non-parallel sides of a trapezium is parallel to the base and is equal to half the sum of the parallel sides.

Solution :

Fig.

We have a trapezium ABCD in which AB||DC. Taking A as origin of reference, let AB→ = b→ and AD→ = d→. Thus position vector of B is b→ and that of D is d→. Now DC→ || AB→ => DC→ = λAB→ = λb→ for some scalar λ ------- (1) Since AC→ = AD→ + DC→ = d→ + λb→
Therefore, position vector of C is d→ + λb→.
If E and F be the mid-points of BC and AD, then position vector of E is 1/2(b→ + d→ + λb→) and that of F is 1/2d→.
Therefore, FE→ = AE→ – AF→
= 1/2(b→ + d→ + λb→) – 1/2d→
= 1/2(1 + λ)b→
=> FE→ is a scalar multiple of b→
=> FE→ || b→
=> FE→ || AB→.
Also, FE→ = 1/2(1 + λ)AB→ -------------------- (2)
Now, AB→ + DC→ = AB→ + λAB→ [by (1)]
= (1 + λ)AB
= 2 FE→ [by (2)]
=> FE→ 1/2(AB→ + DC→). [Proved.]

Paper By Mr. M.P.Keshari
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