A) \[-4\mathbf{a}\]
B) \[4\mathbf{a}-\mathbf{b}-\mathbf{c}\]
C) \[4\mathbf{c}\]
D) None of these
Correct Answer: D
Solution :
The equations of the lines joining \[6\mathbf{a}-4\mathbf{b}+4\mathbf{c},-4\mathbf{c}\] and \[-\mathbf{a}-2\mathbf{b}-3\mathbf{c},\mathbf{a}+2\mathbf{b}-5\mathbf{c}\]are respectively. \[\mathbf{r}=6\mathbf{a}-4\mathbf{b}+4\mathbf{c}+m(-6\mathbf{a}-4\mathbf{b}-8\mathbf{c})\] ?..(i) and \[\mathbf{r}=-\mathbf{a}-2\mathbf{b}-3\mathbf{c}+n(2\mathbf{a}+4\mathbf{b}-2\mathbf{c})\] ?..(ii) For the point of intersection, the equations (i)and (ii) should give the same value of \[\mathbf{r}\]. Hence, equating the coefficients of vectors \[\mathbf{a},\mathbf{b}\] and \[\mathbf{c}\]in the two expressions for \[\mathbf{r}\], we get \[6m+2n=7,\,\,2m-2n=1\] and \[8m-2n=7\]. Solving first two equations, we get \[m=1\], \[n=\frac{1}{2}\]. These values of m and n also satisfy the third equation. Hence, the lines intersect. Putting the value of m in (i), we get the position vector of the point of intersection as \[-4\mathbf{c}\].
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