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Question

The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1) having normal perpendicular to both the lines
x11=y+22=z43 and x22=y+11=z+71, is


A

2074 units

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B

1083 units

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C

583 units

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D

1074 units

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Solution

The correct option is B

1083 units


Given, equations of lines are
x11=y+22=z43 and x22=y+11=z+71
Let n1=^i2^j+3^k and n2=2^i^j^k
Any vector n prependicular to both n1,n2 is given by n=n1×n2
∣ ∣ ∣^i ^j ^k1 2 32 1 1∣ ∣ ∣=5^i+7^j+3^k
Equation of a plane passing through (1, -1, -1) and perpendicular to n is given by
5(x1)+7(y+1)+3(z+1)=0 5x+7y+3z+5=0
Required distance =5+2121+552+72+32=1083 units


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