Science:Math Exam Resources/Courses/MATH105/April 2016/Question 01 (a)

MATH105 April 2016

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Question 01 (a)
Find an equation of the plane that passes through the point $(-1,0,3)$ with a normal vector $\langle 3,5,-2\rangle$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
What is the equation of plane when the normal vector and a point on the plane are given?

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. An equation of a plane that passes through the point $P=(x_{0},y_{0},z_{0})$ and is orthogonal to a vector $n=\langle a,b,c\rangle$ (a normal vector) is given by $a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$ or by the simplified form $ax+by+cz=d,$ where $d=ax_{0}+by_{0}+cz_{0}$ . Here, we have the point $P=(-1,0,3)$ and the normal vector $n=\langle 3,5,-2\rangle$. So, the equation of the plane is $3(x-(-1))+5(y-0)+(-2)(z-3)=0$, which implies $3(x+1)+5y-2z+6=0$ . We then get ${\begin{aligned}3x+5y-2z=-9.\end{aligned}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $(x,y,z)$ be a point on the plane. Then the vector $(x,y,z)-(-1,0,3)=(x+1,y,z-3)$ lies on the plane. Since a normal vector on a plane is orthogonal to any vector lying on the plane, we have ${\begin{aligned}0=(x+1,y,z-3)\cdot (3,5,-2)=3(x+1)+5y-2(z-3)=3x+5y-2z+9\end{aligned}}$. Therefore, the equation for the plane is $3x+5y-2z=-9$.