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CBSE Class 12 Math 2020 Outside Delhi Set 3 Solved Paper

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Question : 3 of 11
Marks: +1, -0
The line   x−23=  y−34=  z−45\;\frac{x-2}{3} = \;\frac{y-3}{4} = \;\frac{z-4}{5} is parallel to the plane
Solution:     👈: Video Solution
Explanation for the correct option:
Given: line   x−23=  y−34=  z−45\;\frac{x-2}{3} = \;\frac{y-3}{4} = \;\frac{z-4}{5}
let   x−23=  y−34=  z−45=λ\;\frac{x-2}{3} = \;\frac{y-3}{4} = \;\frac{z-4}{5} = \lambda
3i^+4j^+5k^=a⃗3\hat{i} + 4\hat{j} + 5\hat{k} = \vec{a} will be perpendicular to normal vector of a plane.
Checking option DD
So,
=3(2)+4(1)+5(−2)= 3(2) + 4(1) + 5(-2)
=6+4−10= 6 + 4 - 10
=0= 0
Hence, 2x+y−2z=02x + y - 2z = 0 is parallel to   x−23=  y−34=  z−45\;\frac{x-2}{3} = \;\frac{y-3}{4} = \;\frac{z-4}{5}.
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