JEE MAIN - Mathematics (2022 - 28th June Morning Shift - No. 15)
If $$\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k$$, $$\overrightarrow b = 3\widehat i + 3\widehat j + \widehat k$$ and $$\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$$ are coplanar vectors and $$\overrightarrow a \,.\,\overrightarrow c = 5$$, $$\overrightarrow b \bot \overrightarrow c $$, then $$122({c_1} + {c_2} + {c_3})$$ is equal to ___________.
Answer
150
Explanation
$$2{C_1} + {C_2} + 3{C_3} = 5$$ ...... (i)
$$3{C_1} + 3{C_2} + {C_3} = 0$$ ...... (ii)
$$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{ 2 & 1 & 3 \cr 3 & 3 & 1 \cr {{C_1}} & {{C_2}} & {{C_3}} \cr } } \right|$$
$$ = 2(3{C_3} - {C_2}) - 1(3{C_3} - {C_1}) + 3(3{C_2} - 3{C_1})$$
$$ = 3{C_3} + 7{C_2} - 8{C_1}$$
$$ \Rightarrow 8{C_1} - 7{C_2} - 3{C_3} = 0$$ ...... (iii)
$${C_1} = {{10} \over {122}},{C_2} = {{ - 85} \over {122}},{C_3} = {{225} \over {122}}$$
So $$122({C_1} + {C_2} + {C_3}) = 150$$
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