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

The correct option is C √(14) Given, | u #160;| =1,| v #160;| =2,| w #160;| =3 #160;The projection of v along #160; u = v ...

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Byju's Answer

Standard XII

Mathematics

Area between Two Curves

Let u⃗,v⃗,w...

Question

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

2

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\sqrt{7}

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\sqrt{14}

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14

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Solution

The correct option is C $\sqrt{14}$
Given, $∣ ∣ ¯ ¯ ¯ u ∣ ∣ = 1, ∣ ∣ ¯ ¯ ¯ v ∣ ∣ = 2, ∣ ∣ ¯ ¯¯ ¯ w ∣ ∣ = 3$
The projection of $¯ ¯ ¯ v$ along $\to u = \frac{\to v . \to u}{∣ ∣ ¯ ¯ ¯ u ∣ ∣}$ and the projection of $\to u$ along $\to w = \frac{\to w . \to u}{∣ ∣ \to u ∣ ∣}$
So, $\frac{\to v . \to u}{∣ ∣ ¯ ¯ ¯ u ∣ ∣} = \frac{\to w . \to u}{∣ ∣ \to u ∣ ∣} \Rightarrow \to v . \to u = \to w . \to u$
and $\to v, \to w$ are perpendicular to each other
$∴ \to v . \to w = 0 {∣ ∣ \to u - \to v + \to w ∣ ∣}^{2} = {∣ ∣ \to u ∣ ∣}^{2} + {∣ ∣ \to v ∣ ∣}^{2} {∣ ∣ \to w ∣ ∣}^{2} - 2 \to u . \to v + 2 \to u . \to w - 2 \to v . \to w$
$\Rightarrow {∣ ∣ \to u - \to v + \to w ∣ ∣}^{2} = 1 + 4 + 9 - 2 \to u . \to v + 2 \to v . \to u$
$\Rightarrow {∣ ∣ \to u - \to v + \to w ∣ ∣}^{2} = 1 = 4 = 9$
$\Rightarrow ∣ ∣ \to u - \to v + \to w ∣ ∣ = \sqrt{14}$

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Similar questions

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

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