If A(1,1,2), B(4,2,1) and C(2,3,5) are the ve | vedclass

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

Question

(B) Let $AD$ be the median through vertex $A$. Since $D$ is the midpoint of $BC$, the coordinates of $D$ are given by:$D = \left(\frac{4+2}{2}, \frac{

Vedclass · Accepted Answer

$(1+2 t) \hat{i}+\left(1+\frac{3 t}{2}\right) \hat{j}+(2+t) \hat{k}$

Vedclass · Answer

(B) Let $AD$ be the median through vertex $A$. Since $D$ is the midpoint of $BC$, the coordinates of $D$ are given by:$D = \left(\frac{4+2}{2}, \frac{2+3}{2}, \frac{1+5}{2}\right) = \left(3, \frac{5}{2}, 3\right)$The vector equation of the line passing through $A(1, 1, 2)$ and $D(3, 5/2, 3)$ is given by $\vec{r} = \vec{a} + t(\vec{d} - \vec{a})$, where $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{d} = 3\hat{i} + \frac{5}{2}\hat{j} + 3\hat{k}$.$\vec{r} = (\hat{i} + \hat{j} + 2\hat{k}) + t((3-1)\hat{i} + (\frac{5}{2}-1)\hat{j} + (3-2)\hat{k})$$\vec{r} = (\hat{i} + \hat{j} + 2\hat{k}) + t(2\hat{i} + \frac{3}{2}\hat{j} + \hat{k})$$\vec{r} = (1+2t)\hat{i} + (1+\frac{3}{2}t)\hat{j} + (2+t)\hat{k}$

If $A(1,1,2)$, $B(4,2,1)$ and $C(2,3,5)$ are the vertices of a triangle, then a vector representing the median of the triangle through $A$ is

Similar Questions

If the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda}$ and $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$ are coplanar,then $\sin ^{-1}(\sin \lambda)+\cos ^{-1}(\cos \lambda)=$

The equation of the line passing through the points $(3, 2, 4)$ and $(4, 5, 2)$ is

The sum of all values of $ \alpha $, for which the shortest distance between the lines $ \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} $ and $ \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} $ is $ \sqrt{2} $, is

If $A(3,4,5)$, $B(4,6,3)$, $C(-1,2,4)$, and $D(1,0,5)$ are points such that the angle between the lines $DC$ and $AB$ is $\theta$, then $\cos \theta$ is equal to

If the harmonic conjugate of $P(2, 3, 4)$ with respect to the line segment joining the points $A(3, -2, 2)$ and $B(6, -17, -4)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module