MHT CET · Maths · Three Dimensional Geometry
The vector equation of the line \(\frac{x+3}{2}=\frac{2 y-3}{5} ; z=-1\) is
- A \(\bar{r}=\left(3 \hat{\imath}-\frac{3}{2} \hat{\jmath}-\hat{k}\right)+\lambda(4 \hat{\imath}+5 \hat{\jmath})\)
- B \(\bar{r}=\left(-3 \hat{\imath}+\frac{3}{2} \hat{\jmath}-\hat{k}\right)+\lambda(4 \hat{\imath}+5 \hat{\jmath})\)
- C \(\bar{r}=\left(-3 \hat{\imath}+\frac{3}{2} \hat{\jmath}+\hat{k}\right)+\lambda(4 \hat{\imath}+5 \hat{\jmath})\)
- D \(\bar{r}=\left(3 \hat{\imath}+\frac{3}{2} \hat{\jmath}-\hat{k}\right)+\lambda\left(4 \hat{\imath}+\frac{5}{2} \hat{\jmath}\right)\)
Answer & Solution
Correct Answer
(B) \(\bar{r}=\left(-3 \hat{\imath}+\frac{3}{2} \hat{\jmath}-\hat{k}\right)+\lambda(4 \hat{\imath}+5 \hat{\jmath})\)
Step-by-step Solution
Detailed explanation
Equation of line is \(\frac{x+3}{2}=\frac{2 y-3}{5} ; z=-1\)
\(\therefore \frac{x+3}{2}=\frac{2\left(y-\frac{3}{2}\right)}{5} ; z=-1 \quad \Rightarrow \frac{x+3}{2}=\frac{y-\frac{3}{2}}{\left(\frac{5}{2}\right)} ; z=-1\)
This line passes through point \(\left(-3, \frac{3}{2},-1\right)\) and d.r.s. are \(2, \frac{5}{2}, 0\) i.e. \(4,5,0\) Hence vector equation of given line is
\(\overline{\mathrm{r}}=\left(-3 \overline{\mathrm{i}}+\frac{3}{2} \hat{\mathrm{j}}-\hat{\mathrm{k}}\right)+\lambda(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}})\)
\(\therefore \frac{x+3}{2}=\frac{2\left(y-\frac{3}{2}\right)}{5} ; z=-1 \quad \Rightarrow \frac{x+3}{2}=\frac{y-\frac{3}{2}}{\left(\frac{5}{2}\right)} ; z=-1\)
This line passes through point \(\left(-3, \frac{3}{2},-1\right)\) and d.r.s. are \(2, \frac{5}{2}, 0\) i.e. \(4,5,0\) Hence vector equation of given line is
\(\overline{\mathrm{r}}=\left(-3 \overline{\mathrm{i}}+\frac{3}{2} \hat{\mathrm{j}}-\hat{\mathrm{k}}\right)+\lambda(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}})\)
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