Question

# Find vector equation of line passing through (3,−1,2) and perpendicular to the lines¯¯¯r=¯i+¯j−¯¯¯k+λ(2¯i−2¯j+2¯¯¯k) and ¯¯¯r=2¯i+¯j−3¯¯¯k+μ(¯i−2¯j+2¯¯¯k)

A
¯¯¯r=4¯i+2¯j2¯¯¯k+λ(2¯i2¯j+2¯¯¯k)
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B
¯¯¯r=¯i+¯j¯¯¯k+λ(2¯i3¯j+4¯¯¯k)
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C
¯¯¯r=3¯i+¯j2¯¯¯k+λ(2¯i2¯j+2¯¯¯k)
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D
¯¯¯r=3¯i¯j+2¯¯¯k+λ(2¯i3¯j2¯¯¯k)
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Solution

## The correct option is C ¯¯¯r=3¯i−¯j+2¯¯¯k+λ(−2¯i−3¯j−2¯¯¯k)the question has error the eq of second line will be→r=2^i+^j−3^k+μ(^i−2^j+4^k)in place of→r=2^i+^j−3^k+μ(^i−2^j+2^k)required line is passing through A(3,-1,2) position vector become OA=→a=3^i−^j+2^kSo eq of line become →r=→a+λ→b-------------(1)required line is perpendicular to lines→r=^i+^j−^k+λ(2^i−2^j+2^k)normal vector of above line →n1=2^i−2^j+2^kand required line is also perpendicular to lines→r=2^i+^j−3^k+μ(^i−2^j+4^k)normal vector of above line→n2=^i−2^j+4^kline is perpendicular to both lines so→b=→n1×→n2→b=∣∣ ∣ ∣∣^i^j^k2−221−24∣∣ ∣ ∣∣→b=^i(−8+4)−^j(8−2)+^k(−4+2)→b=−4^i−6^j−2^kputting a and b in eq (1)So required line of eq →r=3^i−^j+2^k+λ(−4^i−6^j−2^k)→r=3^i−^j+2^k+2λ(−2^i−3^j−^k)→r=3^i−^j+2^k+λ(−2^i−3^j−^k)This is required eq

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