JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 21)
Let $$\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,$$ $$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k$$, $$\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$$ be three vectors such that the projection vector of $$\overrightarrow b $$ on $$\overrightarrow a $$ is $$\overrightarrow a $$.
If $$\overrightarrow a + \overrightarrow b $$ is perpendicular to $$\overrightarrow c $$ , then $$\left| {\overrightarrow b } \right|$$ is equal to :
If $$\overrightarrow a + \overrightarrow b $$ is perpendicular to $$\overrightarrow c $$ , then $$\left| {\overrightarrow b } \right|$$ is equal to :
$$\sqrt {32} $$
6
$$\sqrt {22} $$
4
Explanation
Projection of $$\overrightarrow b $$ on $$\overrightarrow a $$ is $$\overrightarrow a $$
$$ \therefore $$ $${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$$
$$ \Rightarrow $$ $${{{b_1} + {b_2} + 2} \over 2} = 2$$
$$ \Rightarrow $$ $${b_1} + {b_2} = 2\,\,\,\,\,\,\,\,\,\,\,....(1)$$
and $$\overrightarrow a $$ + $$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$
$$ \Rightarrow $$ $$\left( {\overrightarrow a + \overrightarrow b } \right) \cdot \overrightarrow c = 0$$
$$ \Rightarrow $$ $$5\left( {{b_1} + 1} \right) + \left( {{b_2} + 1} \right) + \sqrt 2 \left( {2\sqrt 2 } \right) = 0$$
$$ \Rightarrow $$ $$5{b_1} + {b_2} + 10 = 0\,\,\,\,\,\,\,\,\,\,......(2)$$
solving (1) & (2)
b1 $$=$$ $$-$$ 3 and b2 $$=$$ 5
$$ \Rightarrow $$ $$\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6$$
$$ \therefore $$ $${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$$
$$ \Rightarrow $$ $${{{b_1} + {b_2} + 2} \over 2} = 2$$
$$ \Rightarrow $$ $${b_1} + {b_2} = 2\,\,\,\,\,\,\,\,\,\,\,....(1)$$
and $$\overrightarrow a $$ + $$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$
$$ \Rightarrow $$ $$\left( {\overrightarrow a + \overrightarrow b } \right) \cdot \overrightarrow c = 0$$
$$ \Rightarrow $$ $$5\left( {{b_1} + 1} \right) + \left( {{b_2} + 1} \right) + \sqrt 2 \left( {2\sqrt 2 } \right) = 0$$
$$ \Rightarrow $$ $$5{b_1} + {b_2} + 10 = 0\,\,\,\,\,\,\,\,\,\,......(2)$$
solving (1) & (2)
b1 $$=$$ $$-$$ 3 and b2 $$=$$ 5
$$ \Rightarrow $$ $$\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6$$
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