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QJEEM09Mae7c3: JEE Main 2009 Mathematics - Class 12 JEEM09Mae7c3 Solution: The projections of a vector on the three coordinate axis are…

The projections of a vector on the three coordinate axis are 6,−3,26,-3,2 respectively. The direction cosines of the vector are :

Multiple Choice Options

  • A 65,−35,25{6 \over 5},{{ - 3} \over 5},{2 \over 5}
  • B 67,−37,27{6 \over 7 },{{ - 3} \over 7},{2 \over 7}
  • C −67,−37,27{- 6 \over 7 },{{ - 3} \over 7},{2 \over 7}
  • D 6,−3,26, -3, 2

Step-by-Step Verified Solution

Let P(x1,y1,z1)P\left( {{x_1},{y_1},{z_1}} \right) and Q(x2,y2,z2)Q\left( {{x_2},{y_2},{z_2}} \right) be the initial and final points of the vector whose projections on the three coordinates axes are 6,−3,2{6, - 3,2} then

x2−x1,=6;  y2−y1=−3;  z2−z1=2{x_2} - {x_1}, = 6;\,\,{y_2} - {y_1} = - 3;\,\,{z_2} - {z_1} = 2

So that directions ratios of PQ→\overrightarrow {PQ} are 6,−3,2{6, - 3,2}

∴\therefore Direction cosines of PQ→\overrightarrow {PQ} are

662+(−3)2+22,−362+(−3)2+22,{6 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},{{ - 3} \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},

        \,\,\,\,\,\,\,\, 262+(−3)2+22=67,−37,27{2 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }} = {6 \over 7},{{ - 3} \over 7},{2 \over 7}

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