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 (integer)
150
Solution
<p>$2{C_1} + {C_2} + 3{C_3} = 5$ ...... (i)</p>
<p>$3{C_1} + 3{C_2} + {C_3} = 0$ ...... (ii)</p>
<p>$$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{
2 & 1 & 3 \cr
3 & 3 & 1 \cr
{{C_1}} & {{C_2}} & {{C_3}} \cr
} } \right|$$</p>
<p>$= 2(3{C_3} - {C_2}) - 1(3{C_3} - {C_1}) + 3(3{C_2} - 3{C_1})$</p>
<p>$= 3{C_3} + 7{C_2} - 8{C_1}$</p>
<p>$\Rightarrow 8{C_1} - 7{C_2} - 3{C_3} = 0$ ...... (iii)</p>
<p>$${C_1} = {{10} \over {122}},{C_2} = {{ - 85} \over {122}},{C_3} = {{225} \over {122}}$$</p>
<p>So $122({C_1} + {C_2} + {C_3}) = 150$</p>
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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