Vector Algebra

1 )

if |a| = 3.5 then |a × i|² + |a × j|² + |a × k|² =_______

24.5

26.5

View Solution

Answer :

a = xi + yj + zk

a × i = – yk + zj

∴ |a × i| = √[(– y)² + (z)²] = √(y² + z²)

∴ |a × i|² = y² + z² similarly |a × j|² = z² + x² and |a × k|² = x² + y²

∴ |a × i|² + |a × j|² + |a × k|² = 2|x² + y² + z²|

= 2|a|²

= 2(3.5)² ____ |a| = 3.5

= 24.5

2 )

a is non zero vector which magnitude |a|, m is scalar. If ma is unit vector satisfied ____

Select correct option from foloowing options

(a) m = ± 1

(b) m = |a|

(d) m = ± 2

View Solution

Answer :

ma is unit vector ⇒ |ma| = 1

∴ m|a| = 1

∴ m = ± [1 / (|a|)]

3 )

If 2i + j, i – 3j, 3i + 2j & i + λj are position vectors of points A, B, C, D respectively and AB || CD. Then λ is ____

– 8

– 6

– 4

View Solution

Answer :

AB = (i – 3j) – (2i + j) = – i – 4j

and CD = (i + λj) – (3i + 2j)

i.e. CD = – 2i + (λ – 2)j

Now AB || CD

⇒ [(– 1) / (– 2)] = [(– 4) / (λ – 2)]

∴ λ = – 6

4 )

a and b non coplanar. 2u – v = w, if u = xa + 2yb, v = – 2ya + 3xb and w = 4a – 2b, find x and y = ____

x = (8/7), y = (2/7)

x = 2, y = 3

x = (4/7), y = (6/7)

x = (10 / 7), y = (4/7)

View Solution

Answer :

2u – v = w

2(xa + 2yb) – (– 2ya + 3xb) = 4a – 2b

∴ 2xa + 4yb + 2ya – 3xb = 4a – 2b

∴ 2x + 2y = 4 and 4y – 3x = – 2

∴ x + y = 2 and 4y – 3x = – 2

∴ 4(2 – x) – 3x = – 2

∴ 8 – 7x = – 2

∴ – 7x = – 10

x = (10 / 7) and y = (4/7)

5 )

Two non zero vectors cross product is zero. Then vectors are ____

coplanar

equal vectors

origin at one point

same ending point

View Solution

Answer :

given a × b = 0 ⇒ a & b are parallel vector ⇒ coplanar vector

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