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

Solution

✅ Correct Option: 2

Equal projections on b⃗\vec b mean a⃗⋅b⃗=c⃗⋅b⃗\vec a\cdot\vec b=\vec c\cdot\vec b, and perpendicularity gives a⃗⋅c⃗=0\vec a\cdot\vec c=0. Expanding, ∣2a⃗+3b⃗−2c⃗∣2=4(4)+9(1)+4(9)+12(a⃗⋅b⃗)−12(b⃗⋅c⃗)−8(a⃗⋅c⃗)=61|2\vec a+3\vec b-2\vec c|^2=4(4)+9(1)+4(9)+12(\vec a\cdot\vec b)-12(\vec b\cdot\vec c)-8(\vec a\cdot\vec c)=61, since the middle terms cancel. So the value is 61\sqrt{61}.

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