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Question
A perpendicular is drawn from the point A(0,1,−1) to the line joining the points B(0,−1,3) and C(2,−3,−1). The length of the perpendicular is:
A perpendicular is drawn from the point A(0,1,−1) to the line joining the points B(0,−1,3) and C(2,−3,−1). The length of the perpendicular is:
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Solution
The correct option is B √14
Equation of line BC is : x2=y+1−2=z−3−4
Any point D on the line can be taken as: (2t,−2t−1,−4t+3)
Now, −−→DA=(−2t,2+2t,−4+4t)
For foot of perpendicular, −−→DA.(2^i−2^j−4^k)=0
⇒−4t−4−4t+16−16t=0
⇒t=12
Now, −−→DA=(−1,3,−2)
Hence, |−−→DA|=√1+9+4=√14
Equation of line BC is : x2=y+1−2=z−3−4
Any point D on the line can be taken as: (2t,−2t−1,−4t+3)
Now, −−→DA=(−2t,2+2t,−4+4t)
For foot of perpendicular, −−→DA.(2^i−2^j−4^k)=0
⇒−4t−4−4t+16−16t=0
⇒t=12
Now, −−→DA=(−1,3,−2)
Hence, |−−→DA|=√1+9+4=√14
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