Let the position vectors of points 'A' and 'B' be i+j+k and 2i+j+3k, respectively. A point 'P' divides the line segment AB internally in the ratio λ:1(λ>0). If O is the origin and OB.OP - 3|OA*OP|² = 6, then λ is equal to
Let the position vectors of points 'A' and 'B' be i+j+k and 2i+j+3k, respectively. A point 'P' divides the line segment AB internally in the ratio λ:1(λ>0). If O is the origin and OB.OP - 3|OA*OP|² = 6, then λ is equal to
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|3a+4b|² = 9|a|² + 16|b|² + 24a·b
But a·b = 0, |a|=|b|=k
|3a+4b| = 5k
|4a-3b|
10k = 20? k = 2 = |a| = |b|
Given r × a = r × b, which means r × a - r × b = 0 ⇒ r × (a - b) = 0.
This implies that vector r is parallel to vector (a - b).
So, r = λ (a - b) for some scalar λ.
a - b = (2i - 3j + 4k) - (7i + j - 6k) = -5i - 4j + 10k.
So, r = λ (-5i - 4j + 10k).
We are also given r ⋅ (i + 2j + k) = -3.
λ (-5i - 4j + 1
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