A) \[\vec{r}=\left( \frac{8}{6}\hat{i}+\frac{7}{2}\hat{j} \right)+t\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{3}\hat{k} \right)\]
B) \[\vec{r}=\left( \frac{8}{6}\hat{i}+\frac{7}{2}\hat{j} \right)+s\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{3}\hat{k} \right)\]
C) \[\vec{r}=\left( -\frac{8}{6}\hat{i}-\frac{7}{2}\hat{j} \right)+t\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{3}\hat{k} \right)\]
D) \[\vec{r}=\left( -\frac{8}{6}\hat{i}-\frac{7}{2}\hat{j} \right)+s\left( \frac{1}{6}\hat{i}-\frac{1}{2}\hat{j}-\frac{1}{3}\hat{k} \right)\]
Correct Answer: B
Solution :
Given, equation of straight line is \[6x-8=2y-7=3z\] \[\Rightarrow \] \[6\left( x-\frac{8}{6} \right)=2\left( y-\frac{7}{2} \right)=3\,(z-0)\] \[\Rightarrow \] \[\frac{x-\frac{8}{6}}{\frac{1}{6}}=\frac{y-\frac{7}{2}}{\frac{1}{2}}=\frac{z-0}{1/3}\] Vector equation of this straight line is \[\vec{r}=\left( \frac{8}{6}\hat{i}+\frac{7}{2}\hat{j}+0\hat{k} \right)+s\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{3}\hat{k} \right)\] \[\vec{r}=\left( \frac{8}{6}\hat{i}+\frac{7}{2}\hat{j} \right)+s\left( \frac{1}{6}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{3}\hat{k} \right)\]
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