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Mathematics Question on Three Dimensional Geometry

The vector equation of the straight line 6x−8=2y−7=3z6x-8=2y-7=3z is

A

r⃗=(86i^+72j^)+t(16i^+12j^−13k^)\vec{r}=\left( \frac{8}{6}\hat{i}+\frac{7}{2}\hat{j} \right)+t\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{3}\hat{k} \right)

B

r⃗=(86i^+72j^)+s(16i^+12j^−13k^)\vec{r}=\left( \frac{8}{6}\hat{i}+\frac{7}{2}\hat{j} \right)+s\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{3}\hat{k} \right)

C

r⃗=(−86i^−72j^)+t(16i^+12j^−13k^)\vec{r}=\left( -\frac{8}{6}\hat{i}-\frac{7}{2}\hat{j} \right)+t\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{3}\hat{k} \right)

D

r⃗=(−86i^−72j^)+s(16i^−12j^−13k^)\vec{r}=\left( -\frac{8}{6}\hat{i}-\frac{7}{2}\hat{j} \right)+s\left( \frac{1}{6}\hat{i}-\frac{1}{2}\hat{j}-\frac{1}{3}\hat{k} \right)

Answer

r⃗=(86i^+72j^)+s(16i^+12j^−13k^)\vec{r}=\left( \frac{8}{6}\hat{i}+\frac{7}{2}\hat{j} \right)+s\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{3}\hat{k} \right)

Explanation

Solution

Given, equation of straight line is
6x−8=2y−7=3z6x-8=2y-7=3z
⇒\Rightarrow 6(x−86)=2(y−72)=3 (z−0)6\left( x-\frac{8}{6} \right)=2\left( y-\frac{7}{2} \right)=3\,(z-0)
⇒\Rightarrow x−8616=y−7212=z−01/3\frac{x-\frac{8}{6}}{\frac{1}{6}}=\frac{y-\frac{7}{2}}{\frac{1}{2}}=\frac{z-0}{1/3}
Vector equation of this straight line is
r⃗=(86i^+72j^+0k^)+s(16i^+12j^+13k^)\vec{r}=\left( \frac{8}{6}\hat{i}+\frac{7}{2}\hat{j}+0\hat{k} \right)+s\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{3}\hat{k} \right)
r⃗=(86i^+72j^)+s(16i^+12j^+13k^)\vec{r}=\left( \frac{8}{6}\hat{i}+\frac{7}{2}\hat{j} \right)+s\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{3}\hat{k} \right)