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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $B$.

Similar Questions

The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is:

Consider a $\triangle ABC$ where $A(1,3,2)$,$B(-2,8,0)$,and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$,then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:

If the coordinates of $A, B, C, D$ are $(2, 3, -1), (3, 5, -3), (1, 2, 3)$ and $(3, 5, 7)$ respectively,then what is the projection of $AB$ on $CD$?

If $A, B, C$ and $D$ are four points in the plane such that $|AB|^2+|CD|^2=|BC|^2+|DA|^2=100$,then $AC \cdot BD=$

If $\bar{a}$ and $\bar{b}$ are unit vectors such that $(\bar{a} + 2\bar{b})$ and $(5\bar{a} - 4\bar{b})$ are perpendicular to each other,then the angle between $\bar{a}$ and $\bar{b}$ is .....$^o$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module