Vector Algebra

6 )

x and y are nonzero vector. If x = ky, k < 0 then x ∙ y = ____

Select correct option from following options

(a) = |x + y|

(b) = |x| |y|

(d) < |x| |y|

View Solution

Answer :

x = ky and k is < 0

⇒ x & y as e in opposite direction

⇒ x ∙ y < |x| |y|

7 )

(x ∙ y) ∙ z is what ?

unit vector

vector

scalar

None of these

View Solution

Answer :

(x ∙ y) ∙ z is not possible.

8 )

If a + mb + 3c, – 2a + 3b – 4c and a – 3b – 5c are coplanar. m = ____

â€“ (9/7)

(9/7)

View Solution

Answer :

Vectors are coplanar

⇒ |D| = 0

1	m	3
– 2	3	– 4	= 0
1	– 3	– 5

∴ 1(– 15 – 12) – m(10 + 4) + 3(6 – 3) = 0

∴ – 27 – 14m + 9 = 0

∴ 14m = – 18

m = [(– 18) / (14)] = – (9/7)

9 )

x is nonzero vector find real number k such that |(5 – k) x | < 2|x| is satisfied ____

0 < k < 3

â€“ 7 < k < â€“ 3

3 < k < 7

â€“ 7 < k < 3

View Solution

Answer :

given: |(5 – k)x| < 2|x|

∴ |5 – k| |x| < 2 |x|

also |x| ≠ 0 as x ≠ 0

∴ |5 – k| < 2 ........ dividing by |x|

∴ 5 – k < 2 or – 5 + k < 2

i.e. 3< k or k < 7

i.e. 3 < k < 7

10 )

let u, v and w such that |u| = 1, |v| = 2, |w| = 3. projection of v over u and projection of w over u are same magnitude. v and w is perpendicular | u – v + w | = ____

âˆš14

âˆš7

View Solution

Answer :

|u| = 1, |v| = 2 and |w| = 3, v ∙ w = 0

consider |u – v + w|² = (u – v + w) ∙ (u – v + w)

= |u|² + |v|² + |w|² – 2u ∙ v + 2w ∙ u – 2v ∙ w

= 1² + 2² + 3² – 2(|u| ∙ |v| cos β) + 2(|w| ∙ |u| cos α) – 0

= 14 + 2(|w| |u| cos α – |u| |v| cos β)

= 14 + 0 ...... as projections of same magnitude

∴ |u – v + w| = √14

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