A) \[\frac{11}{7}\]
B) \[\frac{21}{7}\]
C) \[\frac{22}{7}\]
D) \[\frac{17}{7}\]
Correct Answer: C
Solution :
[c] \[\because \]Given that the direction ratio of a line is \[6,2,3.\] \[\therefore \]Direction cosines of the given line be \[\frac{6}{\sqrt{{{6}^{2}}+{{2}^{2}}+{{3}^{2}}}},\frac{2}{\sqrt{49}},\frac{3}{\sqrt{49}}=\frac{6}{7},\frac{2}{7},\frac{3}{7}\] Hence, the projection of the line segment joining \[\left( -1,0,3 \right)\] and \[\left( 2,5,1 \right)\] on the line whose direction cosines are \[\left( \frac{6}{7},\frac{2}{7},\frac{3}{7} \right)\]is \[=1\left( {{x}_{2}}-{{x}_{1}} \right)+m\left( {{y}_{2}}-{{y}_{1}} \right)+n\left( {{z}_{2}}-{{z}_{1}} \right)\] \[=\frac{6}{7}.(2+1)+\frac{2}{7}.(5-0)+\frac{3}{7}(1-3)\] \[=\frac{6\times 3+2\times 5+3\times (-2)}{7}=\frac{18+10-6}{7}=\frac{22}{7}\] Hence, option [c] is correct.
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