QJEEM18Maccea: JEE Main 2018 Physics - Class 11 JEEM18Maccea Solution: Let be a vector coplanar with the vectors and . If is…

Question

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Let

\overrightarrow u

be a vector coplanar with the vectors

\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k

and

\overrightarrow b = \widehat j + \widehat k

. If

\overrightarrow u

is perpendicular to

\overrightarrow a

and

\overrightarrow u .\overrightarrow b = 24

, then

{\left| {\overrightarrow u } \right|^2}

is equal to

Accepted Answer

You should know that, when

\overrightarrow u

is coplanar with

\overrightarrow a

and

\overrightarrow b

then we can write

\overrightarrow u = x\overrightarrow a + y\overrightarrow b

Here,

\overrightarrow u

is perpendicular with

\overrightarrow a

then,

\overrightarrow u .\overrightarrow a = 0

\Rightarrow \,\,\,\,\left( {x\,\overrightarrow a + y\overrightarrow b } \right)\,.\overrightarrow a = 0

\Rightarrow \,\,\,\,x\,.\overrightarrow {\,a} \,.\,\overrightarrow a \, + \,y\,.\,\overrightarrow a \,.\,\overrightarrow b = 0

\Rightarrow \,\,\,\,x\,.\,{\left| {\overrightarrow a } \right|^2} + \,y\overrightarrow a \,.\,\overrightarrow b = 0

[ As

\left| {\overrightarrow a } \right| = \sqrt {{2^2} + {3^2} + {{\left( { - 1} \right)}^2}} = \sqrt {14}

and

\overrightarrow a .\overrightarrow b = \left( {2\widehat i + 3\widehat j - \widehat k} \right).\left( {\widehat j + \widehat k} \right)

= \,\,\,\,\left( {2.0 + 3.1 + \left( { - 1} \right).1} \right)

= \,\,\,\,2

]

\Rightarrow \,\,\,\,x\,.\,\left( {14} \right) + y\,.\,2 = 0

\Rightarrow \,\,\,\,7x\, + \,y\, = 0........\left( 1 \right)

Given,

\overrightarrow u \,.\,\overrightarrow b = 24

\Rightarrow \,\,\,\,\,\left( {x\overrightarrow a + y\overrightarrow b } \right).\overrightarrow b = 24

\Rightarrow \,\,\,\,x\left( {\overrightarrow a .\overrightarrow b } \right) + y{\left| {\overrightarrow b } \right|^2} = 24

\Rightarrow \,\,\,\,x.2 + y.{\left( {\sqrt {{1^2} + {1^2}} } \right)^2} = 24

\Rightarrow \,\,\,\,2x + 2y = 24

\Rightarrow \,\,\,\,x + y = 12.......\left( 2 \right)

By solvig (1) and (2) we get,

x = - 2 and y = 14

Now,

{\left| {\overrightarrow u } \right|^2} = \overrightarrow u .\overrightarrow u

= \,\,\,\,\left( {x\overrightarrow a + y\overrightarrow b } \right).\overrightarrow u

= \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u

= \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u

= \,\,\,\,0 + 14 \times 24

[as

\overrightarrow a .\overrightarrow u = 0

and

\overrightarrow u .\overrightarrow b = 24]

=\,\,\,\,

336

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