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

|u|=1,|v|=2,|w|=3 . By the given condition, ⇒v·u|u|=w·v|u| ⇒v·u=w·u Also v·w=0 (∵v⊥w) Now|u v+w|^2=u^2+v^2+w^2 2 u·v 2 v·w+2 w·u =1+4+9 2 (0) =14 ∴|u v+w| ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Applications Scalar Triple Product

Let u,v, and ...

Question

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{14}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

14

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\sqrt{14}$
$| \to u | = 1, | \to v | = 2, | \to w | = 3 .$ By the given condition,
$\Rightarrow \frac{\to v \cdot \to u}{| \to u |} = \frac{\to w \cdot \to v}{| \to u |}$

$\Rightarrow \to v \cdot \to u = \to w \cdot \to u$

Also $\to v \cdot \to w = 0 (∵ \to v ⊥ \to w)$

Now $| \to u - v + \to w |^{2} = {\to u}^{2} + {\to v}^{2} + {\to w}^{2} - 2 \to u \cdot \to v - 2 \to v \cdot \to w + 2 \to w \cdot \to u$

$= 1 + 4 + 9 - 2 (0) = 14 ∴ | \to u - \to v + \to w | = \sqrt{14}$

Suggest Corrections

Similar questions

Q. Let

u, v

and

w

be such that

| u | = 1, | v | = 3

and

| w | = 2

. If the projection of

v

along

u

is equal to that of

w

along

u

and vectors

v

and

w

are perpendicular to each other, then

| u - v + w |

equals

Q. 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| =

Q. 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 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

Q. 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| =

Q. 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 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

Join BYJU'S Learning Program

Explore more

Applications Scalar Triple Product

Standard XII Mathematics

Join BYJU'S Learning Program

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

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