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

|u¯|=1,|v¯|=2|w¯|=3 v¯.u¯|u¯|=w¯.u¯|u¯|⇒v¯.u¯=w¯.u¯ And v¯⊥w¯⇒v¯.w¯=0 |u¯−v¯+w¯|2=|u¯|2+|v¯|2+|w¯|2−2u¯.v¯+2u¯.w¯−2v¯.w¯ =1+4+9−2(0) =14 ∴|u¯−v¯+w¯|=14

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

|u¯|=1,|v¯|=2|w¯|=3 v¯.u¯|u¯|=w¯.u¯|u¯|⇒v¯.u¯=w¯.u¯ And v¯⊥w¯⇒v¯.w¯=0 |u¯−v¯+w¯|2=|u¯|2+|v¯|2+|w¯|2−2u¯.v¯+2u¯.w¯−2v¯.w¯ =1+4+9−2(0) =14 ∴|u¯−v¯+w¯|=14

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

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$\sqrt{14}$

$\sqrt{7}$

answer is C.

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Detailed Solution

$| \bar{u} | = 1, | \bar{v} | = 2 | \bar{w} | = 3$

$\frac{\bar{v} . \bar{u}}{| \bar{u} |} = \frac{\bar{w} . \bar{u}}{| \bar{u} |} \Rightarrow \bar{v} . \bar{u} = \bar{w} . \bar{u}$

And $\bar{v} ⊥ \bar{w} \Rightarrow \bar{v} . \bar{w} = 0$

${| \bar{u} - \bar{v} + \bar{w} |}^{2} = {| \bar{u} |}^{2} + {| \bar{v} |}^{2} + {| \bar{w} |}^{2} - 2 \bar{u} . \bar{v} + 2 \bar{u} . \bar{w} - 2 \bar{v} . \bar{w}$