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Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3hati+5hatj-6hatk |
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Answer» `therefore` The unit vector normal to the plane `=(1)/abs(3hati+5hatj-6hatk) (3hati+5hatj-6hatk) = (1)/sqrt(9+25+36) (3hati+5hatj-6hatk) = (3hati+5hatj-6hatk)/sqrt70` Hence, the vector EQUATION of the plane is `VECR.((3hati+5hatj-6hatk)/sqrt70) = 7` |
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