If a, b, c are non-coplanar vectors,then the | vedclass

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(B) Let the position vectors of the points be $A = 2a+3b-c$,$B = 3a+4b-2c$,$C = a-2b+3c$,and $D = a-6b+6c$.The vector equation of the line passing thr

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$a+2b$

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(B) Let the position vectors of the points be $A = 2a+3b-c$,$B = 3a+4b-2c$,$C = a-2b+3c$,and $D = a-6b+6c$.The vector equation of the line passing through $A$ and $B$ is given by $r = A + t(B-A)$,where $t \in R$.$r = (2a+3b-c) + t((3a+4b-2c) - (2a+3b-c))$$r = (2a+3b-c) + t(a+b-c) = (2+t)a + (3+t)b + (-1-t)c$ ... $(i)$The vector equation of the line passing through $C$ and $D$ is given by $r = C + s(D-C)$,where $s \in R$.$r = (a-2b+3c) + s((a-6b+6c) - (a-2b+3c))$$r = (a-2b+3c) + s(0a-4b+3c) = (1)a + (-2-4s)b + (3+3s)c$ ... (ii)Since the lines intersect,we equate the coefficients of $a, b, c$ in $(i)$ and (ii) because $a, b, c$ are non-coplanar:For $a$: $2+t = 1 \Rightarrow t = -1$.For $b$: $3+t = -2-4s \Rightarrow 3-1 = -2-4s \Rightarrow 2 = -2-4s \Rightarrow 4s = -4 \Rightarrow s = -1$.Check for $c$: $-1-t = 3+3s \Rightarrow -1-(-1) = 3+3(-1) \Rightarrow 0 = 0$. This is consistent.Substituting $t = -1$ into equation $(i)$:$r = (2-1)a + (3-1)b + (-1-(-1))c = a + 2b + 0c = a+2b$.Thus,the point of intersection is $a+2b$.

If $a, b, c$ are non-coplanar vectors,then the point of intersection of the line passing through the points $2a+3b-c$ and $3a+4b-2c$ with the line joining the points $a-2b+3c$ and $a-6b+6c$ is

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