If the line joining points A and B having pos | vedclass

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(A) Let the point of intersection divide $AB$ in the ratio $\lambda : 1$ and $CD$ in the ratio $\mu : 1$.The position vector of a point on $AB$ is $\v

Vedclass · Accepted Answer

$B$

Vedclass · Answer

(A) Let the point of intersection divide $AB$ in the ratio $\lambda : 1$ and $CD$ in the ratio $\mu : 1$.The position vector of a point on $AB$ is $\vec{r} = \frac{\lambda(-4 \vec{c}) + 1(6 \vec{a}-4 \vec{b}+4 \vec{c})}{\lambda+1} = \frac{6 \vec{a}-4 \vec{b} + (4-4 \lambda) \vec{c}}{\lambda+1}$.The position vector of a point on $CD$ is $\vec{r} = \frac{\mu(\vec{a}+2 \vec{b}-5 \vec{c}) + 1(-\vec{a}-2 \vec{b}-3 \vec{c})}{\mu+1} = \frac{(\mu-1) \vec{a} + (2 \mu-2) \vec{b} + (-5 \mu-3) \vec{c}}{\mu+1}$.Equating the two expressions for $\vec{r}$:$\frac{6 \vec{a}-4 \vec{b} + (4-4 \lambda) \vec{c}}{\lambda+1} = \frac{(\mu-1) \vec{a} + (2 \mu-2) \vec{b} + (-5 \mu-3) \vec{c}}{\mu+1}$.Comparing coefficients of $\vec{a}$ and $\vec{b}$:$\frac{6}{\lambda+1} = \frac{\mu-1}{\mu+1}$ and $\frac{-4}{\lambda+1} = \frac{2 \mu-2}{\mu+1}$.Dividing the two equations: $\frac{6}{-4} = \frac{\mu-1}{2(\mu-1)} \Rightarrow -\frac{3}{2} = \frac{1}{2}$,which implies $\mu=1$.Substituting $\mu=1$ into the first equation: $\frac{6}{\lambda+1} = 0$,which is impossible unless $\lambda 	o \infty$.However,checking the point $B$ (where $\lambda=0$): $\vec{r} = 6 \vec{a}-4 \vec{b}+4 \vec{c}$.Checking the point $C$ (where $\mu=0$): $\vec{r} = -\vec{a}-2 \vec{b}-3 \vec{c}$.Re-evaluating the intersection: The point $B$ corresponds to $\lambda=0$ on $AB$. If we set $\mu=1$ in the $CD$ line equation,we get $\vec{r} = \frac{0 \vec{a} + 0 \vec{b} - 8 \vec{c}}{2} = -4 \vec{c}$,which is exactly point $B$. Thus,the intersection is $B$.

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