Question

Let p be the plane in space that intersects the x-axis at −5, the y-axis at −2, and the z-axis at 5. Find a vector v¯¯¯ that is perpendicular to p.

Answer 1

Let P be the plane that intersects

• x-axis at point (-5,0,0);
• y-axis at point (0,-2,0);
• z-axis at point (0,0,5).

Write the equation of the plane P:

$\left|\begin{array}{ccc}x-(-5)&y-0&z-0\\0-(-5)&-2-0&0-0\\0-(-5)&0-0&5-0\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x+5&y&z\\5&-2&0\\5&0&5\end{array}\right|=0.$

Then

$-10(x+5)+10z-25y=0,\\ \\2(x+5)-2z+5y=0,\\ \\2x+5y-2z=-10.$

The coefficients at variables x, y and z are the coordinates of perpendicular vector to the plane. Thus

$\vec{v}=(2,5,-2)\perp P.$

Answer: $\vec{v}=(2,5,-2).$