AP EAMCET201822 Apr 2018Morning ShiftMathematicsVector AlgebraActual
Let a , b and c be three non-coplanar vectors. The vector equation of a line which passes through the point of intersection of two lines, one joining the points a +2 b -5 c , - a -2 b -3 c and the other joining the points -4 c , 6 a -4 b +4 c is
Options
- Ar=2 a-4 b+3 c+ (a-6 b+4 c)
- Br=3 a+6 b-c+ (a+2 b+c)
- Cr =2 a +3 ~b - c + ( a + b - c )
- Dr=-2 b+3 c+ (a-4 b+3 c)
Correct answer
B. r=3 a+6 b-c+ (a+2 b+c)
Step-by-step solution
Equation of line joining the points, a +2 b -5 c ,- a -2 b -3 c is r =( a +2 b -5 c )+ ₁,(2 a +4 b -2 c ) Similarly, equation of the line joining the points -4 c, 6 a-4 b+4 c is r =-4 c + ₂(6 a -4 b +8 c ) Now, for point of intersection of lines (i) and (ii) aligned (2 ₁+1 ) a + & (4 ₁+2 ) b + (-2 _ , -5 ) c = & (6 ₂ ) a + (-4 ₂ ) b + (8 ₂+4 ) c aligned On comparing aligned 2 ₁+1 & =6 ₂ 4 ₁+2 & =-4 ₂ -2 ₁-5 & =8 ₂+4 aligned and From Eqs. (iii), (iv) and (v) ₁=- 1 2 and ₂=0 So, intersection points is -4 c and for =-