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Chapter 9: Problem 16

Determine the location of a point \((x, y, z)\) that satisfies the condition(s). \(x y<0, \quad z=4\)

Short Answer

Expert verified
The point that satisfies the condition is (2, -3, 4).

Step by step solution

01

Understand the conditions

First, we need to understand what the conditions \(x y<0\) and \(z=4\) tell us. The condition \(xy<0\) means that the product of x and y should be less than zero, that is the number should be negative. Now, the product of two numbers is negative when one number is positive and the other one is negative. As for the condition \(z=4\), it simply means that the third coordinate, z should be equal to 4.
02

Determine the point

21Since the product xy should be less than zero, one of the numbers, x or y, should be negative while the other should be positive. It does not matter which of x or y is positive or negative. The point that fulfills these conditions is (2, -3, 4). Here, x 路 y equals -6, which is less than zero, and z equals 4. So, the point (2, -3, 4) satisfies the required conditions.

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Most popular questions from this chapter

Find the magnitude of \(v\). Initial point of \(\mathbf{v}:(1,-3,4)\) Terminal point of \(\mathbf{v}:(1,0,-1)\)

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 the vector \(z,\) given that \(\mathbf{u}=\langle 1,2,3\rangle\) \(\mathbf{v}=\langle 2,2,-1\rangle,\) and \(\mathbf{w}=\langle 4,0,-4\rangle\) \(2 \mathbf{u}+\mathbf{v}-\mathbf{w}+3 \mathbf{z}=\mathbf{0}\)

In Exercises 61 and \(62,\) use vectors to determine whether the points are collinear. (0,-2,-5),(3,4,4),(2,2,1)

Prove the triangle inequality \(\|\mathbf{u}+\mathbf{v}\| \leq\|\mathbf{u}\|+\|\mathbf{v}\|\).
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