Question

Let `bar"a", bar"b", bar"c"` be three vectors such that `bar"a" ≠ 0`, and `bar"a" xx bar"b" = 2bar"a" xx bar"c", |bar"a"| = |bar"c"| = 1, |bar"b"| = 4` and `|bar"b" xx bar"c"| = sqrt(15)`. If `bar"b" - 2bar"c" = lambdabar"a"`, then λ is equal to ______.

Options

• 1

• ±4

• 3

• –2

Answer 1

Let `bar"a", bar"b", bar"c"` be three vectors such that `bar"a" ≠ 0`, and `bar"a" xx bar"b" = 2bar"a" xx bar"c", |bar"a"| = |bar"c"| = 1, |bar"b"| = 4` and `|bar"b" xx bar"c"| = sqrt(15)`. If `bar"b" - 2bar"c" = lambdabar"a"`, then λ is equal to ±4.

Explanation:

If angle between `bar"b"` and `bar"c"` is `alpha` and `|bar"b" xx bar"c"| = sqrt(15)`

⇒  `|bar"b"| |bar"c"| sin alpha = sqrt(15)`

⇒ `sin alpha = sqrt(15)/4`

⇒ `cos alpha = 1/4`

Now, `bar"b" - 2bar"c" = lambdabar"a"`

⇒ `|bar"b" - 2bar"c"|^2 = lambda^2|bar"b"|^2`

⇒ `|bar"b"|^2 + 4|bar"c"|^2 - 4bar"b"*bar"c" = lambda^2|bar"a"|^2`

⇒ `16 + 4 - 4 {|bar"b"| |bar"c"| cos alpha} = lambda^2`

⇒ `16 + 4 - 4 xx 4 xx 1 xx 1/4 = lambda^2`

⇒ λ² = 16

⇒ λ = ±4