Let \overline{a}, \overline{b}, \overline{c} be coinitial vectors and \overline{a}=2 \overline{i}-\overline{j}+5 \overline{k} and \overline{b}=3 \overline{i}+7 \overline{j}-\overline{k}. Let (\overline{a}, \overline{b})=\theta be an acute angle and \overline{c} be the vector along the bisector of the angle \theta. If \lambda, x, y \in \mathbb{R}, then \overline{c}=

Correct Answer is : (2x+3y)i+(7y−x)j+(5x−y)k

(5 \overline{ {\i}}+6 \overline{ {\j}}+4 \overline{ {\k}})

2 \lambda(-\overline{ {\i}}-8 \overline{ {\j}}+6 \overline{ {\k}})

(2 x+3 y) \overline{i}+(7 y-x) \overline{j}+(5 x-y) \overline{ {\k}}

(2 x+3 y) \overline{i}+(x+7 y) \overline{ {\j}}+(5 x+y) \overline{ {\k}}

AP EAPCET 18th May 2023 Shift 2 Solved Paper

Section: Mathematics

Question : 30 of 160

Marks: +1, -0

Let a,b,c be coinitial vectors and a=2i−j+5k and b=3i+7j−k. Let (a,b)=θ be an acute angle and c be the vector along the bisector of the angle θ. If λ,x,y∈ℝ, then c=

(5i+6j+4k)
2λ(−i−8j+6k)
(2x+3y)i+(7y−x)j+(5x−y)k
(2x+3y)i+(x+7y)j+(5x+y)k

Validate

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