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How do you find the angle between the planes 2x+5y-z=6 and 3x-2y+6z=10?

Answer 1

The angle is #105.1º#

You look for the angle between the vectors normal to the planes: Plane1, the vector is#〈2,5,-1〉# Plane2, the vector is #〈3,-2,6〉# The dot product is #〈2,5,-1〉.〈3,-2,6〉=6-10-6=-10# The modulus of vector1 is #sqrt(4+25+1)=sqrt30# The modulus of vector2 is #sqrt(9+4+36)=sqrt49=7# The #costheta=-10/(7sqrt30)=-0.26# #theta # is the angle between the planes #:. theta=105.1º#

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Answer 2

Ezekiel Alexander

To find the angle between the planes (2x + 5y - z = 6) and (3x - 2y + 6z = 10), you can use the formula:

[ \cos \theta = \frac{{\text{{normal vector of plane 1}} \cdot \text{{normal vector of plane 2}}}}{{\lVert \text{{normal vector of plane 1}} \rVert \cdot \lVert \text{{normal vector of plane 2}} \rVert}} ]

Find the normal vectors of the planes.
Calculate the dot product of the normal vectors.
Use the formula to find the angle ( \theta ).

After finding ( \theta ), you can use ( \cos^{-1} ) to find the angle in radians or degrees.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.