Consider the plane that contains points A(2,3,1), B(-11,1,2), C(-7,-3,-6).
a) Find two vectors that are parallel to the plane.
Ans: AC, BC or AB will be parallel to the plane.
b) Find two vectors that are perpendicular to the plane.
Ans: if i find the normal vector then i can use dot product and use arbitrary values to find the two vectors
c) Write a vector equation of the plane
Ans: I have two points AC, BC or AB and i have the normal, so i use them to write the vector equation.
d) Write the scalar equation of the plane.
Ans: the normal vectors are my coefficients for the(x,y,z) therefore if i substitute a point to find the constant 'd'
e) Write an equation of the line through the x- and y- intercepts of the plane
Ans: Find the x and y intercepts and those will be my points my z value being zero and using the normal as direction vector
Are the answers correct? I am a little confused, please correct if i am wrong somewhere.
Your understanding is mostly correct, but there are a few points that need clarification.
a) To find two vectors that are parallel to the plane, you can subtract the coordinates of any two points on the plane. So, in this case, you can find vectors AC, BC, or AB.
b) To find two vectors that are perpendicular to the plane, you need the normal vector of the plane. To find the normal vector, you can use the cross product of vectors AB and AC. The cross product will give you a vector that is perpendicular to both AB and AC, hence perpendicular to the plane.
c) The vector equation of a plane can be written as:
r = a + s(u) + t(v)
where r is a position vector in the plane, a is a known point on the plane, u and v are vectors parallel to the plane, and s and t are scalars. In this case, you can use point A as the known point and vectors AC and AB as the vectors parallel to the plane.
So, the vector equation of the plane would be:
r = A + s(AC) + t(AB)
d) The scalar equation of a plane can be written as:
ax + by + cz = d
where a, b, and c are the coefficients of the normal vector, and d is a constant. To find the scalar equation, you can use the normal vector and substitute the coordinates of a point on the plane, such as point A, into the equation.
e) To find the equation of the line through the x- and y-intercepts of the plane, you would need to find the x and y intercepts first. The intercepts can be found by setting z = 0 in the scalar equation and solving for x and y. Once you have the intercepts, you can write the equation of the line using the intercepts as points and the normal vector as the direction vector.
So, overall, your approach is correct, but there are some additional steps and clarifications needed.