JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 9)
Let the position vectors of points 'A' and 'B' be
$$\widehat i + \widehat j + \widehat k$$ and $$2\widehat i + \widehat j + 3\widehat k$$, respectively. A point 'P' divides the line segment AB internally in the ratio $$\lambda $$ : 1 ( $$\lambda $$ > 0). If O is the origin and
$$\overrightarrow {OB} .\overrightarrow {OP} - 3{\left| {\overrightarrow {OA} \times \overrightarrow {OP} } \right|^2} = 6$$, then $$\lambda $$ is equal to______.
$$\widehat i + \widehat j + \widehat k$$ and $$2\widehat i + \widehat j + 3\widehat k$$, respectively. A point 'P' divides the line segment AB internally in the ratio $$\lambda $$ : 1 ( $$\lambda $$ > 0). If O is the origin and
$$\overrightarrow {OB} .\overrightarrow {OP} - 3{\left| {\overrightarrow {OA} \times \overrightarrow {OP} } \right|^2} = 6$$, then $$\lambda $$ is equal to______.
Answer
0.8
Explanation
Let, $$\overrightarrow a $$ = $$\widehat i + \widehat j + \widehat k$$
and $$\overrightarrow b $$ = $$2\widehat i + \widehat j + 3\widehat k$$_2nd_September_Evening_Slot_en_9_2.png)
$$\overrightarrow {OB} = \overrightarrow b $$
$$\overrightarrow {OP} = {{\overrightarrow a + \lambda \overrightarrow b } \over {1 + \lambda }}$$
$$\overrightarrow {OA} = \overrightarrow a $$
$$ \therefore $$ $$\overrightarrow {OB} .\overrightarrow {OP} - 3{\left| {\overrightarrow {OA} \times \overrightarrow {OP} } \right|^2} = 6$$
$$ \Rightarrow $$ $$\overrightarrow b .{{\left( {\overrightarrow a + \lambda \overrightarrow b } \right)} \over {1 + \lambda }} - 3\left| {\overrightarrow a \times {{\left( {\overrightarrow a + \lambda \overrightarrow b } \right)} \over {1 + \lambda }}} \right|$$ = 6
$$ \Rightarrow $$ $${{\overrightarrow b .\overrightarrow a + \lambda \left( {\overrightarrow b .\overrightarrow b } \right)} \over {1 + \lambda }} - 3\left| {{{\overrightarrow a \times \overrightarrow a + \lambda \left( {\overrightarrow a \times \overrightarrow b } \right)} \over {1 + \lambda }}} \right|$$ = 6
[ $${\overrightarrow b .\overrightarrow a }$$ = ($$2\widehat i + \widehat j + 3\widehat k$$).($$\widehat i + \widehat j + \widehat k$$)
= 2 + 1 + 3 = 6
$${\overrightarrow b .\overrightarrow b }$$ = ($$2\widehat i + \widehat j + 3\widehat k$$).($$2\widehat i + \widehat j + 3\widehat k$$)
= 4 + 1 + 9 = 14
$$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 1 & 1 \cr 2 & 1 & 3 \cr } } \right|$$
= (3 - 1)$${\widehat i}$$ - (3 - 2)$${\widehat j}$$ + (1 - 2)$${\widehat k}$$
= 2$${\widehat i}$$ - $${\widehat j}$$ - $${\widehat k}$$
$$ \therefore $$ $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = $$\sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} $$ = $$\sqrt 6 $$]
$$ \Rightarrow $$ $${{6 + 14\lambda } \over {1 + \lambda }} - 3{\left| {{{\lambda \left( {\sqrt 6 } \right)} \over {1 + \lambda }}} \right|^2}$$ = 6
$$ \Rightarrow $$ $${{6 + 14\lambda } \over {1 + \lambda }} - {{3{\lambda ^2} \times 6} \over {{{\left( {1 + \lambda } \right)}^2}}}$$ = 6
$$ \Rightarrow $$ (14$$\lambda $$ + 6)($$\lambda $$ + 1)—18$$\lambda $$2 = 6($$\lambda $$ + 1)2
$$ \Rightarrow $$ —4$$\lambda $$2 + 20$$\lambda $$ + 6 = 6$$\lambda $$2 + 12$$\lambda $$ + 6
$$ \Rightarrow $$ 10$$\lambda $$2 — 8$$\lambda $$ = 0
$$ \Rightarrow $$ $$\lambda $$(10$$\lambda $$ — 8) = 0
As given $$\lambda $$ > 0
$$ \therefore $$ $$\lambda $$ = $${8 \over {10}}$$ = 0.8
and $$\overrightarrow b $$ = $$2\widehat i + \widehat j + 3\widehat k$$
$$\overrightarrow {OB} = \overrightarrow b $$
$$\overrightarrow {OP} = {{\overrightarrow a + \lambda \overrightarrow b } \over {1 + \lambda }}$$
$$\overrightarrow {OA} = \overrightarrow a $$
$$ \therefore $$ $$\overrightarrow {OB} .\overrightarrow {OP} - 3{\left| {\overrightarrow {OA} \times \overrightarrow {OP} } \right|^2} = 6$$
$$ \Rightarrow $$ $$\overrightarrow b .{{\left( {\overrightarrow a + \lambda \overrightarrow b } \right)} \over {1 + \lambda }} - 3\left| {\overrightarrow a \times {{\left( {\overrightarrow a + \lambda \overrightarrow b } \right)} \over {1 + \lambda }}} \right|$$ = 6
$$ \Rightarrow $$ $${{\overrightarrow b .\overrightarrow a + \lambda \left( {\overrightarrow b .\overrightarrow b } \right)} \over {1 + \lambda }} - 3\left| {{{\overrightarrow a \times \overrightarrow a + \lambda \left( {\overrightarrow a \times \overrightarrow b } \right)} \over {1 + \lambda }}} \right|$$ = 6
[ $${\overrightarrow b .\overrightarrow a }$$ = ($$2\widehat i + \widehat j + 3\widehat k$$).($$\widehat i + \widehat j + \widehat k$$)
= 2 + 1 + 3 = 6
$${\overrightarrow b .\overrightarrow b }$$ = ($$2\widehat i + \widehat j + 3\widehat k$$).($$2\widehat i + \widehat j + 3\widehat k$$)
= 4 + 1 + 9 = 14
$$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 1 & 1 \cr 2 & 1 & 3 \cr } } \right|$$
= (3 - 1)$${\widehat i}$$ - (3 - 2)$${\widehat j}$$ + (1 - 2)$${\widehat k}$$
= 2$${\widehat i}$$ - $${\widehat j}$$ - $${\widehat k}$$
$$ \therefore $$ $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = $$\sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} $$ = $$\sqrt 6 $$]
$$ \Rightarrow $$ $${{6 + 14\lambda } \over {1 + \lambda }} - 3{\left| {{{\lambda \left( {\sqrt 6 } \right)} \over {1 + \lambda }}} \right|^2}$$ = 6
$$ \Rightarrow $$ $${{6 + 14\lambda } \over {1 + \lambda }} - {{3{\lambda ^2} \times 6} \over {{{\left( {1 + \lambda } \right)}^2}}}$$ = 6
$$ \Rightarrow $$ (14$$\lambda $$ + 6)($$\lambda $$ + 1)—18$$\lambda $$2 = 6($$\lambda $$ + 1)2
$$ \Rightarrow $$ —4$$\lambda $$2 + 20$$\lambda $$ + 6 = 6$$\lambda $$2 + 12$$\lambda $$ + 6
$$ \Rightarrow $$ 10$$\lambda $$2 — 8$$\lambda $$ = 0
$$ \Rightarrow $$ $$\lambda $$(10$$\lambda $$ — 8) = 0
As given $$\lambda $$ > 0
$$ \therefore $$ $$\lambda $$ = $${8 \over {10}}$$ = 0.8
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