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Determine whether the given vectors are perpendicular. $$ \mathbf{u}=2 \mathbf{i}-8 \mathbf{j}, \quad \mathbf{v}=-12 \mathbf{i}-3 \mathbf{j} $$

Short Answer

Expert verified
The vectors are perpendicular.

Step by step solution

01

Recall the Perpendicular Condition

Two vectors are perpendicular if their dot product is zero. For vectors \( \mathbf{u} \) and \( \mathbf{v} \), the dot product is given by:\[ \mathbf{u} \cdot \mathbf{v} = u_1\cdot v_1 + u_2\cdot v_2 \] where \( u_1 \) and \( u_2 \) are the components of \( \mathbf{u} \), and \( v_1 \) and \( v_2 \) are the components of \( \mathbf{v} \).
02

Identify Components of the Vectors

Identify the components from each vector:\( \mathbf{u} = 2 \mathbf{i} - 8 \mathbf{j} \) is equivalent to components \( u_1 = 2 \) and \( u_2 = -8 \).\( \mathbf{v} = -12 \mathbf{i} - 3 \mathbf{j} \) is equivalent to components \( v_1 = -12 \) and \( v_2 = -3 \).
03

Compute the Dot Product

Substitute the components into the dot product formula:\(\mathbf{u} \cdot \mathbf{v} = (2 \cdot -12) + (-8 \cdot -3)\)Calculate each part:- \( 2 \cdot -12 = -24 \)- \( -8 \cdot -3 = 24 \)Add these results: \(-24 + 24 = 0\).
04

Conclude on Perpendicularity

Since the dot product \( \mathbf{u} \cdot \mathbf{v} = 0 \), vectors \( \mathbf{u} \) and \( \mathbf{v} \) are perpendicular as per the perpendicularity condition.

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Key Concepts

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

Dot Product
In vector mathematics, the dot product is a valuable tool to determine the angle between two vectors. The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is calculated by multiplying their corresponding components and adding the results. The formula in two-dimensional space is:
  • \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 \)
If the dot product equals zero, the vectors are perpendicular, meaning they meet at a right angle. This property simplifies checking perpendicularity in physics and engineering tasks. Don't forget that only vectors in the same dimensional space can have a dot product. It's an essential concept when working with forces, trajectories, and more.
Vector Components
Vectors are quantities with both magnitude and direction. These can be broken down into components along the axes of a coordinate system, which helps in simplifying calculations. For a vector \( \mathbf{u} \) in 2D, with components \( 2 \mathbf{i} \) and \( -8 \mathbf{j} \), \( u_1 = 2 \) and \( u_2 = -8 \).
  • The \( \mathbf{i} \) component indicates the horizontal (x-axis) value.
  • The \( \mathbf{j} \) component indicates the vertical (y-axis) value.
Breaking vectors into components is crucial for calculating magnitudes, angles, and other properties. It allows for intuitive handling of projections and interactions in mechanics, like applying forces.
Vectors in 2D
Vectors in 2D (two dimensions) act within a plane defined by two perpendicular axes, typically the x and y axes. In this plane, a vector can be represented as a combination of its x-component and y-component.
  • For example, the vector \( \mathbf{v} = -12 \mathbf{i} - 3 \mathbf{j} \), consists of an x-component of -12 and a y-component of -3.
Handling vectors in 2D allows for easier visualization and computation. They can represent anything from velocity in a physics problem to forces acting on a structural element. This representation forms a basis for understanding more complex systems and can be extended to three dimensions for more intricate scenarios.

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Most popular questions from this chapter

A description of a line is given. Find parametric equations for the line. The line crosses the \(x\) -axis where \(x=-2\) and crosses the \(z\) -axis where \(z=10\) .

Same Line: Different Parametric Equations Every line can be described by infinitely many different sets of parametric equations, since any point on the line and any vector parallel to the line can be used to construct the equations. But how can we tell whether two sets of parametric equations rep- resent the same line? Consider the following two sets of para- metric equations: Line \(1 : \quad x=1-t, \quad y=3 t, \quad z=-6+5 t\) Line \(2 : \quad x=-1+2 t, \quad y=6-6 t, \quad z=4-10 t\) (a) Find two points that lie on Line 1 by setting \(t=0\) and \(t=1\) in its parametric equations. Then show that these points also lie on Line 2 by finding two values of the parameter that give these points when substituted into the parametric equations for Line 2 . (b) Show that the following two lines are not the same by finding a point on Line 3 and then showing that it does not lie on Line \(4 .\) Line \(3 : \quad x=4 t, \quad y=3-6 t, \quad z=-5+2 t\) Line \(4 : \quad x=8-2 t, \quad y=-9+3 t, \quad z=6-t\)

Find parametric equations for the line that passes through the point \(P\) and is parallel to the vector \(\mathbf{v} .\) $$ P(0,-5,3), \quad \mathbf{v}=\langle 2,0,-4\rangle $$

Find parametric equations for the line that passes through the point \(P\) and is parallel to the vector \(\mathbf{v} .\) $$ P(0,0,0), \quad \mathbf{v}=\langle- 4,3,5\rangle $$

Find the indicated quantity, assuming \(\mathbf{u}=2 \mathbf{i}+\mathbf{j}, \mathbf{v}=\mathbf{i}-3 \mathbf{j},\) and \(\mathbf{w}=3 \mathbf{i}+4 \mathbf{j}\) $$ (\mathbf{u} \cdot \mathbf{v})(\mathbf{u} \cdot \mathbf{w}) $$

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