Question

The points 𝐴(3,−1 ,2) ,𝐵(1 ,2 ,−4 ) and 𝐶 (−1 ,1 ,−2 ) are three vertices of a parallelogram ABCD.

(i) Show that the vector 𝑟 = −16𝑗 – 8𝑘 is perpendicular to the plan through 𝐴 ,𝐵 and 𝐶 .
(6 marks)
(ii) Find the cartesian equation of the plane through 𝐴 ,𝐵 and 𝐶 . (4 marks)

Answer 1

[AI-generated answer — Explain Bot]

To show that the vector 𝑟 = −16𝑗 – 8𝑘 is perpendicular to the plane through points 𝐴, 𝐵, and 𝐶, we can use the dot product of 𝑟 and the normal vector of the plane.

(i) The normal vector of the plane through 𝐴, 𝐵, and 𝐶 can be found by taking the cross product of two vectors formed by subtracting 𝐴 from 𝐵 and 𝐶, respectively.

Let's calculate the normal vector:

Vector 𝐵𝐴 = 𝐵 - 𝐴
= (1, 2, -4) - (3, -1, 2)
= (-2, 3, -6)

Vector 𝐶𝐴 = 𝐶 - 𝐴
= (-1, 1, -2) - (3, -1, 2)
= (-4, 2, -4)

Now, we can calculate the normal vector by taking the cross product of 𝐵𝐴 and 𝐶𝐴:

Normal vector = 𝐵𝐴 x 𝐶𝐴
= (-2, 3, -6) x (-4, 2, -4)

To find the cross product, we can use the formula:

𝑎 = (𝑎_2𝑏_3−𝑎_3𝑏_2)𝑖+(𝑎_3𝑏_1−𝑎_1𝑏_3)𝑗+(𝑎_1𝑏_2−𝑎_2𝑏_1)𝑘

Calculating the cross product:

𝑎 = (3 x (-4) - (-6) x 2)𝑖 + ((-6) x (-4) - (-2) x (-4))𝑗 + ((-2) x 2 - 3 x (-4))𝑘
= (-20)𝑖 + (-8)𝑗 + (2)𝑘
= (-20, -8, 2)

Now we can calculate the dot product of the normal vector and 𝑟:

𝑟 ⋅ Normal vector = (-16𝑗 - 8𝑘) ⋅ (-20, -8, 2)
= 16(20) + 8(8) + 2(0)
= 320 + 64 + 0
= 384

Since the dot product is not equal to zero, 𝑟 is not perpendicular to the plane through 𝐴, 𝐵, and 𝐶.

(ii) The cartesian equation of a plane can be found using the formula:
Ax + By + Cz + D = 0

To find the coefficients A, B, C, and D, we can use the coordinates of one of the points on the plane, for example point 𝐴(3, -1, 2).

Substitute the coordinates of 𝐴 into the equation:

A(3) + B(-1) + C(2) + D = 0

Now, substitute the coefficients A, B, C, and solve for D:

3A - B + 2C + D = 0

We can use the normal vector (-20, -8, 2) as the coefficients A, B, and C:

-20A - 8B + 2C + D = 0

Since the normal vector (-20, -8, 2) is a unique set of coefficients, we can conclude that the cartesian equation of the plane through 𝐴, 𝐵, and 𝐶 is:

-20x - 8y + 2z + D = 0

where D is the constant needed to satisfy the equation based on the coordinates of 𝐴.