JEE MAIN - Mathematics (2017 (Offline) - No. 10)
Let $$\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k$$ and $$\overrightarrow b = \widehat i + \widehat j$$.
Let $$\overrightarrow c $$ be a vector such that $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$,
$$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3$$ and the angle between $$\overrightarrow c $$ and $\overrightarrow a \times \overrightarrow b$ is $$30^\circ $$.
Then $$\overrightarrow a .\overrightarrow c $$ is equal to :
Let $$\overrightarrow c $$ be a vector such that $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$,
$$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3$$ and the angle between $$\overrightarrow c $$ and $\overrightarrow a \times \overrightarrow b$ is $$30^\circ $$.
Then $$\overrightarrow a .\overrightarrow c $$ is equal to :
2
5
$${1 \over 8}$$
$${{25} \over 8}$$
Explanation
Given:
$$\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k,\,\,\overrightarrow b = \widehat i + \widehat j$$
$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right| = 3$$
$$ \therefore $$ $$\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k$$
$$\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{2^2} + {2^2} + {1^2}} = 3$$
We have $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \left| {\overrightarrow a \times \overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin 30^\circ$$
$$ \Rightarrow $$ $$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3\left| {\overrightarrow c } \right|.{1 \over 2}$$
$$ \Rightarrow $$ $$3 = 3\left| {\overrightarrow c } \right|.{1 \over 2}$$
$$ \therefore $$ $$\left| {\overrightarrow c } \right| = 2$$
Now $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$
On squaring, we get
$$ \Rightarrow $$ $${c^2} + {a^2} - 2 - \overrightarrow c .\overrightarrow a = 9$$
$$ \Rightarrow $$ $$4 + 9 - 2 - \overrightarrow a .\overrightarrow c = 9$$
$$ \Rightarrow $$ $$\overrightarrow a .\overrightarrow c = 2$$ [$$ \because $$ $$\overrightarrow c .\overrightarrow a \,\, = \,\,\overrightarrow a .\overrightarrow c $$]
$$\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k,\,\,\overrightarrow b = \widehat i + \widehat j$$
$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right| = 3$$
$$ \therefore $$ $$\overrightarrow a \times \overrightarrow b = 2\widehat i - 2\widehat j + \widehat k$$
$$\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{2^2} + {2^2} + {1^2}} = 3$$
We have $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \left| {\overrightarrow a \times \overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin 30^\circ$$
$$ \Rightarrow $$ $$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right| = 3\left| {\overrightarrow c } \right|.{1 \over 2}$$
$$ \Rightarrow $$ $$3 = 3\left| {\overrightarrow c } \right|.{1 \over 2}$$
$$ \therefore $$ $$\left| {\overrightarrow c } \right| = 2$$
Now $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$
On squaring, we get
$$ \Rightarrow $$ $${c^2} + {a^2} - 2 - \overrightarrow c .\overrightarrow a = 9$$
$$ \Rightarrow $$ $$4 + 9 - 2 - \overrightarrow a .\overrightarrow c = 9$$
$$ \Rightarrow $$ $$\overrightarrow a .\overrightarrow c = 2$$ [$$ \because $$ $$\overrightarrow c .\overrightarrow a \,\, = \,\,\overrightarrow a .\overrightarrow c $$]
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