A) \[x+2y-2z=0\]
B) \[x-2y+z=0\]
C) \[5x+2y-4z=0\]
D) \[3x+2y-3z=0\]
Correct Answer: B
Solution :
| Vector along the normal to the plane containing the lines \[\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\] and \[\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\] is \[(8\hat{i}-j-10\hat{k}).\] |
| Vector perpendicular to the vectors \[2\hat{i}+3\hat{j}+4\hat{k}\]and \[8\hat{i}-\hat{j}-10\hat{k}\]is \[26\hat{i}-52\hat{j}+26\hat{k}\] |
| So, required plane is \[26x-52y+26z=0\] |
| \[\Rightarrow \]\[x-2y+z=0\] |
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