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Chapter 4: Problem 68

Show that the line joining \((2,3,4)\) to \((1,2,3)\) is perpendicular to the line joining \((1,0,2)\) to \((2,3,-2)\)

Short Answer

Expert verified
The lines are perpendicular, as their direction vectors' dot product is 0.

Step by step solution

01

Determine the direction vectors

Find the direction vectors of the lines. For the first line joining \( (2,3,4) \) to \( (1,2,3) \), the direction vector is \((1 - 2, 2 - 3, 3 - 4) = (-1, -1, -1)\). For the second line joining \( (1,0,2) \) to \( (2,3,-2) \), the direction vector is \((2 - 1, 3 - 0, -2 - 2) = (1, 3, -4)\).
02

Use the dot product to check perpendicularity

To determine if the lines are perpendicular, calculate the dot product of their direction vectors. Compute \((-1, -1, -1) \cdot (1, 3, -4) = (-1)(1) + (-1)(3) + (-1)(-4) = -1 - 3 + 4 = 0\).
03

Analyze the result of the dot product

Since the dot product is zero, \(0\), the direction vectors are orthogonal, which means the lines are perpendicular.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direction Vectors
Direction vectors are essential in understanding the orientation of a line in space. Imagine you're moving from one point to another, the path you trace can be represented by a direction vector.
For the direction vector:
  • Start with two points, such as "(2, 3, 4)" and "(1, 2, 3)".
  • Subtract the second point from the first point coordinate-wise.
  • This gives the direction vector "(-1, -1, -1)".
This process is repeated for the second set of points "(1, 0, 2)" to "(2, 3, -2)", yielding the direction vector "(1, 3, -4)". Knowing direction vectors helps visualize the alignment of lines and figure out their relationships.
Dot Product
The dot product is a simple yet powerful tool in vector mathematics that helps us find relationships between vectors. If you take two vectors and multiply corresponding components together and then sum those products, you get the dot product.
Consider two vectors:
  • "\((-1, -1, -1)\)" and "\((1, 3, -4)\)".
Multiply each corresponding component:
  • \((-1) \times 1 = -1\)
  • \((-1) \times 3 = -3\)
  • \((-1) \times (-4) = 4\)
Add these results together:
  • \(-1 + (-3) + 4 = 0\)
The dot product is zero, which provides significant insight into the relationship between the two vectors.
Orthogonality
Orthogonality is a special property in vector mathematics that indicates whether two lines are perpendicular to each other. When the dot product of two vectors is zero, these vectors are said to be orthogonal. Orthogonality has practical applications in many fields, including engineering, physics, and computer graphics.
If two direction vectors are orthogonal:
  • The lines they represent intersect at a 90-degree angle, meaning they are perpendicular.
  • This concept is crucial when you want to prove the perpendicularity of two lines using their direction vectors.
In the exercise, since the dot product of the direction vectors "\((-1, -1, -1)\)" and "\((1, 3, -4)\)" is zero, we conclude that the lines represented by these vectors are indeed perpendicular.

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Most popular questions from this chapter

\(\mathrm{P}\) is a point on a straight line with position vector \(\boldsymbol{r}=\boldsymbol{a}+t \boldsymbol{b}\). Show that $$ r^{2}=a^{2}+2 \boldsymbol{a} \cdot \boldsymbol{b} t+b^{2} t^{2} $$ By completing the square, show that \(r^{2}\) is a minimum for the point \(\mathrm{P}\) for which \(t=-\boldsymbol{a} \cdot \boldsymbol{b} / b^{2}\). Show that at this point \(\overrightarrow{\mathrm{OP}}\) is perpendicular to the line \(\boldsymbol{r}=\boldsymbol{a}+\boldsymbol{t b}\). (This proves the well-known result that the shortest distance from a point to a line is the length of the perpendicular from that point to the line.)

Find the equation of the line joining \((1,-1,3)\) to \((3,3,-1)\). Show that it is perpendicular to the plane \(2 x+4 y-4 z=5\), and find the angle that the line makes with the plane \(12 x-15 y+16 z=10\).

A rigid body is rotating with angular velocity \(6 \mathrm{rad} \mathrm{s}^{-1}\) about an axis in the direction of the vector \((3,-2,1)\) and passing through the point \(\mathrm{A}(3,-2,5)\). Find the linear velocity of the point \(\mathrm{P}(3,-2,1)\) on the body.

If \(\left(\boldsymbol{c}-\frac{1}{2} \boldsymbol{a}\right) \cdot \boldsymbol{a}=\left(\boldsymbol{c}-\frac{1}{2} \boldsymbol{b}\right) \cdot \boldsymbol{b}=0\), prove that the vector \(\boldsymbol{c}-\frac{1}{2}(\boldsymbol{a}+\boldsymbol{b})\) is perpendicular to \(\boldsymbol{a}-\boldsymbol{b}\).

If the position vectors of the points \(\mathrm{P}\) and \(\mathrm{Q}\) are \(i+3 j-7 k\) and \(5 \boldsymbol{i}-2 \boldsymbol{j}+4 \boldsymbol{k}\) respectively, find \(\overrightarrow{\mathrm{PQ}}\) and determine its length and direction cosines.
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