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  • 5 years ago

Determine whether the planes are parallel, perpendicular or neither. If neither find the angle between them. x+4y-3z=1, -3x-12y+6z=1

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  1. anonymous
    • 5 years ago
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    The normal vectors are \[n_1=(1;4;-3),\quad n_2=(-3;-12;6).\] The dot product of n_1 and n_2 \[n_1\cdot n_2=1\cdot(-3)+4\cdot(-12)+(-3)\cdot 6=-69.\] The cosine of the angle between them is \[\cos\theta=\frac{n_1\cdot n_2}{|n_1|\cdot|n_2|}=\frac{-69}{\sqrt{26}\sqrt{189}}\approx -0.984309\] So \[\theta\approx 169.84^\circ\]

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