MHT CET · Maths · Three Dimensional Geometry
The position vector of the point of intersection of the line
\(\bar{r}=(2 \hat{\imath}+\hat{\jmath}-4 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}+2 \hat{k})\) and XOY-Plane is
- A \(4 \hat{\imath}+3 \hat{k}\)
- B \(4 \hat{\imath}+3 \hat{\jmath}\)
- C \(4 \hat{\imath}-3 \hat{k}\)
- D \(4 \hat{\imath}-3 \hat{\jmath}\)
Answer & Solution
Correct Answer
(D) \(4 \hat{\imath}-3 \hat{\jmath}\)
Step-by-step Solution
Detailed explanation
We have line \(\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})\)
Hence coordinates of any point on this line are \((\lambda+2,-2 \lambda+1,2 \lambda-4)\)
This point lies on XOY plane whose equation is \(z=0\) \(\therefore \quad 2 \lambda-4=0 \Rightarrow \lambda=2\)
Hence point of intersection is \((4,-3,0)\). Thus \(\mathrm{pv}\) is \(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}\)
Hence coordinates of any point on this line are \((\lambda+2,-2 \lambda+1,2 \lambda-4)\)
This point lies on XOY plane whose equation is \(z=0\) \(\therefore \quad 2 \lambda-4=0 \Rightarrow \lambda=2\)
Hence point of intersection is \((4,-3,0)\). Thus \(\mathrm{pv}\) is \(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}\)
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