Vector Algebra Question 101
Question: The line joining the points $ 6\mathbf{a}-4\mathbf{b}+4\mathbf{c},,-4\mathbf{c} $ and the line joining the points $ -\mathbf{a}-2\mathbf{b}-3\mathbf{c},,\mathbf{a}+2\mathbf{b}-5\mathbf{c} $ intersect at
Options:
A) $ -4\mathbf{a} $
B) $ 4\mathbf{a}-\mathbf{b}-\mathbf{c} $
C) $ 4\mathbf{c} $
D) None of these
Show Answer
Answer:
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+4n=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} $ .
BETA
