A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
\[a+b+2c=0\] \[2a+3b+4c=0\]\[\Rightarrow \]\[\frac{a}{-2}=\frac{b}{0}=\frac{c}{1}\] For a point on pout z = 0 \[\Rightarrow \] \[x+y=3\] \[2x+3y=4\] By solving we get \[x=5,y=-2,z=0\] \[\therefore \]Point is \[(5,-2,0)\] equation line is\[\frac{x-5}{-2}=\frac{y+2}{0}=\frac{z}{1}\] Shortest distance\[=\left| \frac{({{\overline{a}}_{2}}-{{\overline{a}}_{1}})\times \overline{b}}{\left| \overline{b} \right|} \right|=2\]
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