The shortest distance between the line x=y=z and the line of intersection of 2x+y+z-1=0 and 3x+y+2z-2=0 is

Question

MARKS Content Team · Accepted Answer

Correct answer: (A) 1/(√2) units. P_1=2x+y+z-1=0 P_2=3x+y+2z-2=0 The line of intersection is parallel to i^ j^ k^ 2 1 1 3 1 2 =i^-j^-k^ For the point on the line, putting z=0, we get, 2x+y=1 3x+y=2 x=1, y=-1 So, the required point is 1,-1,0 Hence, the equation of the line of intersection is (x-1)/1=(y+1)/(-1)=(z-0)/(-1) Let, a→=i^-j^ b→=i^-j^-k^ c→=0→ d→=i^-j^-k^ c→-a→=-i^+j^ b→×d→=-2j^+2k^ Shortest distance =(c→-a→⋅b→×d→)/(b→×d→) =(-2)/(2√2)=1/(√2)

The shortest distance between the line $x = y = z$ and the line of intersection of $2 x + y + z - 1 = 0$ and $3 x + y + 2 z - 2 = 0$ is

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