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Let a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} are three vectors such that ∣a⃗∣=3,∣b⃗∣=2,∣c⃗∣=1|\vec{a}|=3, |\vec{b}|=2, |\vec{c}|=1. If the projection of b⃗\vec{b} along c⃗\vec{c} is equal to the projection of a⃗\vec{a} along c⃗\vec{c} and b⃗⊥a⃗\vec{b} \perp \vec{a} , then the value of ∣2a⃗−2b⃗+3c⃗∣|2\vec{a}-2\vec{b}+3\vec{c}| is equal to

Solution

✅ Correct Option: 1

Equal projections along c⃗\vec{c} give a⃗⋅c⃗=b⃗⋅c⃗\vec{a}\cdot\vec{c} = \vec{b}\cdot\vec{c}, and perpendicularity gives a⃗⋅b⃗=0\vec{a}\cdot\vec{b}=0.

Expanding, ∣2a⃗−2b⃗+3c⃗∣2=4(9)+4(4)+9(1)−8(a⃗⋅b⃗)+12(a⃗⋅c⃗−b⃗⋅c⃗)=36+16+9=61|2\vec{a}-2\vec{b}+3\vec{c}|^2 = 4(9)+4(4)+9(1)-8(\vec{a}\cdot\vec{b})+12(\vec{a}\cdot\vec{c}-\vec{b}\cdot\vec{c}) = 36+16+9 = 61.

Hence the magnitude is 61\sqrt{61}.

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