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Consider the equation E_(1):vecr xx (2hati-hatj+3hatk)=3hati+hatk and E_(2): vecr xx (hati+2hatj-3hatk)=2hati-hatj, hten |
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Answer» `E_(1)` represents a line `rArr 3hati(3y+z)-hatj(3x-2z)+hatk(-x-2y)=3hati+hatk` `therefore 3y+z=3, 3x-2y=0, -x-2y=1` From first two EQUATIONS, `3x-2(3-3y)=0` `rArr 3x+6y=6 rArr x+2y=2` Now, `x+2y=-1, x+2y=2` are parallel planes. `E_(2): vecr xx (hati+2hatj-3hatk)=2hati-hatj` `rArr hati(-3y-2z)-hatj(-3x-z)+hatk(2x-y)=2hati-hatj` `therefore hati(-3y-2z)-hatj(-3x-z)+hatk(2x-y)=2hati-hatj` `therefore -3y-2z=2, 3x+z=-1, 2x-y=0` i.e., `-6x-2z=2, 3x+z=-1` `therefore` Straight line, `2x-y=0, 3x+z=-1` |
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