I need to find the equation of a plane that passes through the point with position vector A=5i+j+3k and the line l_1=2i−8j−17k+lambda(i−3j−4k).

Sanai Ball

Sanai Ball

Answered question

2022-09-07

I need to find the equation of a plane that passes through the point with position vector A = 5 i + j + 3 k and the line l 1 = 2 i 8 j 17 k + λ ( i 3 j 4 k )
I found the normal by finding the cross product of A [ ( 5 i + j + 3 k ) ( 2 i 8 j 17 k ) ] which turned out to be n 1 = 24 i 32 j + 18 k. Then I used r n = a n ( 5 i + j + 3 k ) ( 24 i 32 j + 18 k ) = 98
However, it turns out that the above value should be 49 rather than −98. I realize that it can be divided by -2 to get the answer; however, I don't understand why this happens.

Answer & Explanation

farbhas3t

farbhas3t

Beginner2022-09-08Added 6 answers

−98 is correct if you use n 1 = 24 i 32 j + 18 k as normal vector. Then the equation of the plane is
(1) 24 x 32 y + 18 z = 98 .
But you can divide the normal vector by −2, obtaining n 2 = 12 i + 16 j 9 k. For such normal vector the correct value is 49 and the equation of the plane is
(2) 12 x + 16 y 9 z = 49 .
Of course the equations (1) and (2) are equivalent and they represents the same plane.

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