11hey
FUNDA1
first find the equation of the plane 4m the 3 points given and use the formula to get the perp dis
OR
FUNDA2
find the vector AB X AC(unit). The perpendicular distance will be the component of AP on this unit vector
Find the distance between the point P(6, 5, 9) and the plane determined
by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6).
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8 Answers
11
1dude mene aise hi kia tha , lekin mera answer 6/√34 aa ra hai. lekin answer 3√34/17 dia hai
11bhai maine kaha tha unit vector AB X AC did u take the unit vector???
1dude mene first method se kia ............. first waale se b to aana chaiye
11Plane is:
3x-4y+3z=19
hence, in vector form....... r.(3i - 4j + 3k) = 19
distance d= \left|\frac{18-20+27-19}{\sqrt{3^{2}+(-4)^{2}+3^{2}}} \right|
which turns out to be \frac{3\sqrt{2}}{\sqrt{17}}
Hope i am right....
1sorry friends mera answer aur book ka answer same hai.
mera dimaag khraab ho gaya tha