JEE MAIN - Mathematics (2018 (Offline) - No. 10)
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
336
315
256
84
Explanation
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
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
Comments (0)
