Find the distance between the point \( (1,-3,-4) \) and the plane \( 8 y+9 z=0 \). \( d= \) Ex:5.67

To find the distance \( d \) from the point \( (1, -3, -4) \) to the plane defined by the equation \( 8y + 9z = 0 \), we can use the formula for the distance from a point \( (x_0, y_0, z_0) \) to a plane given by the equation \( Ax + By + Cz + D = 0 \): \[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \] 1. First, rewrite the plane equation: \( 8y + 9z = 0 \) can be expressed as \( 0x + 8y + 9z + 0 = 0 \). Thus, \( A = 0 \), \( B = 8 \), \( C = 9 \), and \( D = 0 \). 2. Substitute the point coordinates \( (x_0, y_0, z_0) = (1, -3, -4) \) into the formula: \[ d = \frac{|0(1) + 8(-3) + 9(-4) + 0|}{\sqrt{0^2 + 8^2 + 9^2}} \] 3. Calculate the numerator: \[ = |0 - 24 - 36| = |-60| = 60 \] 4. Calculate the denominator: \[ = \sqrt{0 + 64 + 81} = \sqrt{145} \] 5. Combine to find the distance: \[ d = \frac{60}{\sqrt{145}} \] 6. For a more numerical answer, calculate \( \sqrt{145} \approx 12.04 \): \[ d \approx \frac{60}{12.04} \approx 4.98 \] Thus, the distance \( d \) from the point \( (1, -3, -4) \) to the plane \( 8y + 9z = 0 \) is approximately \( 4.98 \).