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Chapter 9: Problem 4

Find the coordinates of the point. The point is located in the \(y z\) -plane, three units to the right of the \(x z\) -plane, and two units above the \(x y\) -plane.

Short Answer

Expert verified
Therefore, the coordinates of the point are (0, 3, 2).

Step by step solution

01

Determine the x-coordinate

The point is said to be on the yz-plane. In a three-dimensional coordinate system, any point on the yz-plane has an x-coordinate of 0. This is because they lie on the plane perpendicular to the x-axis, that is, without any displacement in the x-direction. Therefore, x = 0.
02

Determine the y-coordinate

The problem states that the point is three units to the right of the xz-plane. In a 3D coordinate system, moving to the right of the xz-plane refers to positive displacement in the y-direction. Therefore, considering the right indicator for positive direction and 3 units for the distance, the y-coordinate equals to 3.
03

Determine the z-coordinate

Finally, the point is said to be two units above the xy-plane. Any movement upwards from the xy-plane is a positive movement in the z-direction. Given that the point is two units above, the z-coordinate equals 2.

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Most popular questions from this chapter

State the definition of parallel vectors.

Use vectors to show that the points form the vertices of a parallelogram. (1,1,3),(9,-1,-2),(11,2,-9),(3,4,-4)

Find \((\mathbf{a}) \mathbf{u} \cdot \mathbf{v},(\mathbf{b}) \mathbf{u} \cdot \mathbf{u},(\mathbf{c})\|\mathbf{u}\|^{2},(\mathbf{d})(\mathbf{u} \cdot \mathbf{v}) \mathbf{v}\) and \((e) u \cdot(2 v)\). $$ \mathbf{u}=\mathbf{i}, \quad \mathbf{v}=\mathbf{i} $$

Find the angle \(\theta\) between the vectors. $$ \begin{array}{l} \mathbf{u}=3 \mathbf{i}+4 \mathbf{j} \\ \mathbf{v}=-2 \mathbf{j}+3 \mathbf{k} \end{array} $$

A point in the three-dimensional coordinate system has coordinates \(\left(x_{0}, y_{0}, z_{0}\right) .\) Describe what each coordinate mea- sures
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