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

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

Question

(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

Vedclass · Accepted Answer

$a+2b$

Vedclass · Answer

(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

Similar Questions

If $\vec{a}$ and $\vec{b}$ are unit vectors such that the vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$,then find the angle between $\vec{a}$ and $\vec{b}$.

If $l\vec{a} + m\vec{b} + n\vec{c} = \vec{0},$ where $l, m, n$ are scalars and $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular non-zero vectors,then

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 2$ and $|\vec{b}| = 3$, then the maximum value of $3 |(3\vec{a} + 2\vec{b})| + 4 |(3\vec{a} - 2\vec{b})|$ is:

The angle between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ is

In $\triangle ABC$,if $\alpha, \beta$ and $\gamma$ are the position vectors of the vertices $A, B$ and $C$ respectively,then the length of the perpendicular from $A$ to $BC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module