Let u-v-w be such that -u-1-v-2-w-3- If the projection of v along u is equal to the projection of w along u and v-w are perpendicular to each other- then -u-v-w-

The correct option is D √(14) Projection of nbsp; v along u =u.v|u|=u.v Projection of w along u =w.u v⊥w⇒v.w=0 and u.v= ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

let u,v,w b...

Question

let $\to u, \to v, \to w$ be such that $∣ ∣ \to u ∣ ∣ = 1, ∣ ∣ \to v ∣ ∣ = 2, ∣ ∣ \to w ∣ ∣ = 3$ . If the projection of $\to v$ along $\to u$ is equal to the projection of $\to w$ along $\to u$ and $\to v, \to w$ are perpendicular to each other, then $∣ ∣ \to u - \to v + \to w ∣ ∣ =$

2

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\sqrt{17}

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\sqrt{14}

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\sqrt{15}

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Solution

The correct option is D $\sqrt{14}$
Projection of $\to v$ along $\to u = \frac{\to u . \to v}{∣ ∣ \to u ∣ ∣} = \to u . \to v$
Projection of $\to w$ along $\to u = \to w . \to u$
$\to v ⊥ \to w \Rightarrow \to v . \to w = 0$ and $\to u . \to v = \to u . \to w (g i v e n)$
Now, ${∣ ∣ \to u - \to v + \to w ∣ ∣}^{2} = {∣ ∣ \to u ∣ ∣}^{2} + {∣ ∣ \to v ∣ ∣}^{2} + {∣ ∣ \to w ∣ ∣}^{2} - 2 \to u . \to v - 2 \to v . \to w + 2 \to u . \to w$
$= 1^{2} + 2^{2} + 3^{2} = 14$ on simplification
$∴ ∣ ∣ \to u - \to v + \to w ∣ ∣ = \sqrt{14}$

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let →u,→v,→w be such that ∣∣→u∣∣=1,∣∣→v∣∣=2,∣∣→w∣∣=3. If the projection of →v along →u is equal to the projection of →w along →u and →v,→w are perpendicular to each other, then∣∣→u−→v+→w∣∣=