3i+j+k,2i+k,i+5j are the position vectors of three non collinear points A,B,C respectively. If the perpendicular drawn from C onto AB meets AB at the point ai+bj+ck, then a+b+c=
So, the direction vector of lin‌E‌A‌B is d‌=−
∧
i
−
∧
j
‌ let ‌P‌=A+λd ‌=(3
∧
i
+
∧
j
+
∧
k
)+λ(−
∧
i
−
∧
j
) ‌=(3−λ)
∧
i
+(1−λ)
∧
j
+
∧
k
and PC=C−P=(
∧
i
+5
∧
j
) ‌−[(3−λ)
∧
i
+(1−λ)
∧
j
+
∧
k
) ‌‌‌=(λ−2
∧
i
+(4+λ)
∧
j
−
∧
k
As, given in the question ‌∵PC⟂AB ‌⇒PC⋅AB=0 ‌⇒(λ−2)(−1)+(4+λ)(−1)+(−1)(0)=0 ‌⇒−λ+2−4−λ=0 ‌⇒λ=−1 ‌∴‌‌P=(3−λ)