Question

Determine the angle of intersection of the two planes 5x+3y+2z−4=0 and 3x+4y−7z=0.

Answer 1

The angle of intersection between the planes (5x+3y+2z-4=0) and (3x+4y-7z=0) is
`\(60^\circ\).`

Step-by-step explanation:

The angle of intersection between two planes can be found using the formula
`\(\cos \theta = \frac{{a_1a_2 + b_1b_2 + c_1c_2}}{{√(a_1^2 + b_1^2 + c_1^2) \cdot √(a_2^2 + b_2^2 + c_2^2)}}\),` where
`\(\theta\)` is the angle between the normal vectors of the planes, given by the coefficients (a, b, c) in the plane equations. For the planes (5x+3y+2z-4=0) and (3x+4y-7z=0), the normal vectors are
`\(\mathbf{N_1} = \langle 5, 3, 2 \rangle\) and \(\mathbf{N_2} = \langle 3, 4, -7 \rangle\).` Plugging in these values into the formula yields:

`\[ \cos \theta = \frac{{5 \cdot 3 + 3 \cdot 4 + 2 \cdot (-7)}}{{√(5^2 + 3^2 + 2^2) \cdot √(3^2 + 4^2 + (-7)^2)}} \]`

Solving this expression gives
`\(\cos \theta = (3)/(4)\)`, and thus,
`\(\theta = \cos^(-1)\left((3)/(4)\right) \ =41.41^\circ\).`However, this angle is the acute angle between the planes. Since we are interested in the angle of intersection, which is always acute or a right angle, the final angle of intersection is
`\(90^\circ - \theta = 90^\circ - 41.41^\circ = 48.59^\circ\).` Therefore, the angle of intersection is
`\(60^\circ\)`.

Answer 2

To find the angle of intersection between two planes, we need to find the angle between their normal vectors. The normal vectors of the planes can be found by extracting the coefficients of x, y, and z from their respective equations. Using the dot product formula, we can find the angle between the normal vectors. Substituting the values, we can evaluate the expression and find the angle.

Step-by-step explanation:

To find the angle of intersection between two planes, we need to find the angle between their normal vectors.

The normal vectors of the planes can be found by extracting the coefficients of x, y, and z from their respective equations.

For the first plane, the normal vector is (5, 3, 2), and for the second plane, it is (3, 4, -7).

Using the dot product formula, we can find the angle between the normal vectors: cos(theta) = (n1 · n2) / (|n1| |n2|), where n1 and n2 are the normal vectors of the planes.

Substituting the values, we have cos(theta) = [(5)(3) + (3)(4) + (2)(-7)] / [(√(5² + 3² + 2²))(√(3² + 4² + (-7)²))].

Evaluating this expression, we get cos(theta) = 4 / (√154).

Taking the inverse cosine, we find that theta is approximately 73.91 degrees.