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The vertices of a triangle ABC are A (1,-2, 2), B (1, 4, 0) and C (-4, 1, 1) respectively. If M be the foot of perpendicular drawn from B on AC, then `vec BM` is |
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Answer» `tantheta=|(m_1-m_2)/(1+m_1m_2)|` `tantheta=|((-3/4)+1)/(1+3/4)|` `tantheta=1/7` `tantheta=|(m+3/4)/(1-3/4m)|` `1/7=|(4m+3)/(4-3m)|` `1/7=(4m+3)/(4-3m)` `m=-17/31,-1` `3x+4y=5` `x+y=1` `3(1-y)+4y=5` `3-3y+4y=5` `3+y=5` `y=2,x=-1` `y-2=-17/31(x+1)` `31y+17x-45=0` `31y+17x=45` option b is correct. |
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