Find the equation of the line passing through (1, 2, 3) and perpendicular to the planes x − y + z = 1 and 2x + y − 3z = 4. A. (x−1)/2 = (y−2)/5 = (z−3)/3 B. (x−1)/2 = (y−2)/−5 = (z−3)/3 C. (x−1)/1 = (y−2)/−1 = (z−3)/1 D. (x−1)/2 = (y−2)/1 = (z−3)/−3 Answer: A) (x−1)/2 = (y−2)/5 = (z−3)/3 Explanation: Direction of line is cross product of normal to the planes: n₁ = (1, −1, 1), n₂ = (2, 1, −3). n₁ × n₂ = i(3−1) − j(−3−2) + k(1+2) = 2i + 5j + 3k. DRs are 2, 5, 3. Equation is (x−1)/2 = (y−2)/5 = (z−3)/3.

imo class 12 three dimensional geometry

Find the equation of the line passing through (1, 2, 3) and perpendicular to the planes x − y + z = 1 and 2x + y − 3z = 4.

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Find the equation of the line passing through (1, 2, 3) and perpendicular to the planes x − y + z = 1 and 2x + y − 3z = 4.

A. (x−1)/2 = (y−2)/5 = (z−3)/3
B. (x−1)/2 = (y−2)/−5 = (z−3)/3
C. (x−1)/1 = (y−2)/−1 = (z−3)/1
D. (x−1)/2 = (y−2)/1 = (z−3)/−3

Answer: A) (x−1)/2 = (y−2)/5 = (z−3)/3

Explanation: Direction of line is cross product of normal to the planes: n₁ = (1, −1, 1), n₂ = (2, 1, −3). n₁ × n₂ = i(3−1) − j(−3−2) + k(1+2) = 2i + 5j + 3k. DRs are 2, 5, 3. Equation is (x−1)/2 = (y−2)/5 = (z−3)/3.

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