Question

|a|=|b|=2,p=a+b,q=a-b,if |p×q|=2[k-(a.b)^2]^1/2 then k =

Answer 1

It has given that, |a| = |b| = 2 , P = a + b , q = a - b

To find : |p × q| = 2[ k - (a.b)² ]½, the value of k = ..

solution : here, p = a + b and q = a - b

so, p × q = (a + b) × (a - b)

= a × a - a × b + b × a - b × b

but we know, a × a = 0 , b × b = 0 [ because sin0° = 0]

so, p × q = -a × b + b × a = 2 (b × a) [as here - a × b = -(-b × a) = b × a ]

now, |p × q| = |2(b × a)| = 2|a||b|sinθ, where θ is angle between a and b.

= 2 × 2 × 2 sinθ

= 8 sinθ

= 2√(4sinθ)²

= 2√(16sin²θ)

we know, sin²θ = 1 - cos²θ

= 2√{16(1 - cos²θ)}

= 2√{16 - 16cos²θ}

= 2√{16 - (4cosθ)²}

now a.b = |a||b|cosθ = 2 × 2cosθ = 4cosθ

= 2√{16 - (a.b)²}

= 2[ 16 - (a.b)² ]½

but |p × q| = [k - (a.b)² ]½

on comparing we get, k = 16

Therefore the value of k = 16

Answer 2

Answer:

k= 16 ....................