QJEEM04Mafc29: JEE Main 2004 Physics - Class 11 JEEM04Mafc29 Solution: Let be such that If the projection along is equal to that of…

Question

Verified Solution • 2025–26 Session moderate

Let

\overrightarrow u ,\overrightarrow v ,\overrightarrow w

be such that

\left| {\overrightarrow u } \right| = 1,\,\,\,\left| {\overrightarrow v } \right|2,\,\,\,\left| {\overrightarrow w } \right|3.

If the projection

{\overrightarrow v }

along

{\overrightarrow u }

is equal to that of

{\overrightarrow w }

along

{\overrightarrow u }

and

{\overrightarrow v },

{\overrightarrow w }

are perpendicular to each other then

\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right|

equals :

Accepted Answer

Projection of

\overrightarrow v

along

\overrightarrow u = {{\overrightarrow v .\overrightarrow u } \over {\left| {\overrightarrow u } \right|}} = {{\overrightarrow v .\overrightarrow u } \over 2}

projection of

\overrightarrow w

along

\overrightarrow u = {{\overrightarrow w .\overrightarrow u } \over {\left| {\overrightarrow u } \right|}} = {{\overrightarrow w .\overrightarrow u } \over 2}

Given

{{\overrightarrow v .\overrightarrow u } \over 2} = {{\overrightarrow w .\overrightarrow u } \over 2}\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

Also,

\overrightarrow v .\overrightarrow w = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

Now

{\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right|^2}

= {\left| {\overrightarrow u } \right|^2} + {\left| {\overrightarrow v } \right|^2} + {\left| {\overrightarrow w } \right|^2} -

\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\overrightarrow u .\overrightarrow v - 2\overrightarrow v .\overrightarrow w + 2\overrightarrow u .\overrightarrow w

= 1 + 4 + 9 + 0

[ From

(1)

and

(2)

]

=14

\therefore

\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right| = \sqrt {14}

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