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Chapter 6: Problem 40

Explain why there does not exist a number \(t\) such that the vectors \((2, t)\) and \((3, t)\) are perpendicular.

Short Answer

Expert verified
Using the dot product, we find that the dot product of the vectors \((2, t)\) and \((3, t)\) is \(6 + t^2\). The vectors are perpendicular if their dot product is zero, which means \(6 + t^2 = 0\). However, there is no real value of \(t\) that satisfies \(t^2 = -6\), because the square of any real number is non-negative. Therefore, there does not exist a number \(t\) such that the vectors \((2, t)\) and \((3, t)\) are perpendicular.

Step by step solution

01

Write down the dot product of the two vectors

The dot product of two vectors \((a, b)\) and \((c, d)\) is given by \(a \cdot c + b \cdot d\). In our case, the two vectors are \((2, t)\) and \((3, t)\). So, their dot product is \(2 \cdot 3 + t \cdot t\).
02

Simplify the dot product

Simplifying the expression for the dot product, we get \[2 \cdot 3 + t \cdot t = 6 + t^2\]
03

Determine the condition for perpendicular vectors

The two vectors are perpendicular if their dot product is zero. Therefore, we want to see if there exists a value of \(t\) such that \[6 + t^2 = 0\]
04

Solve for \(t\)

To find the value of \(t\), we need to solve the equation \[t^2 = -6\] Notice that the left-hand side of this equation is the square of \(t\), while the right-hand side is a negative number.
05

Explain why there is no solution for \(t\)

Since the square of any real number is non-negative, it is impossible to find a real value of \(t\) that satisfies the equation \(t^2 = -6\). Therefore, there is no number \(t\) such that the vectors \((2, t)\) and \((3, t)\) are perpendicular.

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