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Chapter 3: Problem 3

If \(\mathbf{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\) and \(\mathbf{B}=3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\), then vector perpendicular to both \(\mathbf{A}\) and \(\mathbf{B}\) has magnitude \(k\) times that of \((6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})\). That \(k\) is equal to (a) 1 (b) 4 (c) 7 (d) 9

Short Answer

Expert verified
The value of \(k\) is 7, corresponding to option (c).

Step by step solution

01

Find the Cross Product

To find a vector perpendicular to both \(\mathbf{A}\) and \(\mathbf{B}\), we calculate the cross product \(\mathbf{A} \times \mathbf{B}\). Using the cross product formula, we have: \[\mathbf{A} \times \mathbf{B} = \begin{vmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \2 & 3 & 6 \3 & -6 & 2 \end{vmatrix}\]Calculating the determinant, we have:\[\mathbf{A} \times \mathbf{B} = \hat{\mathbf{i}}(3 \cdot 2 - 6 \cdot (-6)) - \hat{\mathbf{j}}(2 \cdot 2 - 6 \cdot 3) + \hat{\mathbf{k}}(2 \cdot (-6) - 3 \cdot 3)\]\[= \hat{\mathbf{i}}[6 + 36] - \hat{\mathbf{j}}[4 - 18] + \hat{\mathbf{k}}[-12 - 9]\]\[= 42 \hat{\mathbf{i}} + 14 \hat{\mathbf{j}} - 21 \hat{\mathbf{k}}\]
02

Calculate the Magnitude of the Cross Product

To determine the magnitude of the vector \(42 \hat{\mathbf{i}} + 14 \hat{\mathbf{j}} - 21 \hat{\mathbf{k}}\), use the formula for the magnitude of a vector: \[\sqrt{42^2 + 14^2 + (-21)^2}\]Calculating inside the root, we have:\[= \sqrt{1764 + 196 + 441}\]\[= \sqrt{2401} = 49\]
03

Calculate the Magnitude of the Given Vector

Find the magnitude of the vector \((6 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 2 \hat{\mathbf{k}})\) using the magnitude formula: \[\sqrt{6^2 + 2^2 + (-2)^2}\]Calculating inside the root:\[= \sqrt{36 + 4 + 4}\]\[= \sqrt{44}\]
04

Find the Value of k

We know the magnitude of the vector from the cross product is \(49\), and the magnitude of \((6 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 2 \hat{\mathbf{k}})\) is \(\sqrt{44}\). According to the problem, \(49 = k \times \sqrt{44}\).To find \(k\), solve:\[k = \frac{49}{\sqrt{44}}\]Calculating, approximate \(\sqrt{44} \approx 6.63\), thus:\[k = \frac{49}{6.63} \approx 7.39\]Hence, \(k\) is approximately \(7\).
05

Compare and Select the Closest Value

We concluded that \(k\) is approximately \(7\). Looking at the options provided, the one that best matches this value is option (c) 7.

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Key Concepts

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

Vector Magnitude
The magnitude of a vector is a measure of its size, without regard to its direction. In simple terms, it tells you how long the vector is. If you visualize a vector as an arrow, the magnitude is the length of this arrow. When calculating it, we use the formula:
  • For a vector \( extbf{V} = a \hat{\textbf{i}} + b \hat{\textbf{j}} + c \hat{\textbf{k}}\), the magnitude, \(|\textbf{V}|\), is \(\sqrt{a^2 + b^2 + c^2}\).
Let's see this in action. For the vector \(\textbf{V} = 42 \hat{\mathbf{i}} + 14 \hat{\mathbf{j}} - 21 \hat{\mathbf{k}}\), the magnitude is:\[\sqrt{42^2 + 14^2 + (-21)^2} = \sqrt{1764 + 196 + 441} = \sqrt{2401} = 49\]Breaking it down further:
  • Square each component: 42 becomes 1764, 14 becomes 196, and -21 becomes 441.
  • Add them together: 1764 + 196 + 441 equals 2401.
  • Finally, take the square root of 2401, which is 49. So, the magnitude is 49.
This tells us the length of our new vector is 49, indicating its size is significant in space.
Perpendicular Vectors
Perpendicular vectors must meet at a right angle of 90 degrees or form a line that looks like an 'L' shape. In vector algebra, two vectors are perpendicular if their dot product equals zero. However, to find a vector perpendicular to both given vectors, we use another concept: the cross product. This results in a new vector.Here's a practical example:
  • If \(\textbf{A} = 2 \hat{\textbf{i}} + 3 \hat{\textbf{j}} + 6 \hat{\textbf{k}}\) and \(\textbf{B} = 3 \hat{\textbf{i}} - 6 \hat{\textbf{j}} + 2 \hat{\textbf{k}}\), a vector perpendicular to both \(\textbf{A}\) and \(\textbf{B}\) is found by calculating \(\textbf{A} \times \textbf{B}\).
  • The result, a new vector, is \(42 \hat{\mathbf{i}} + 14 \hat{\mathbf{j}} - 21 \hat{\mathbf{k}}\), which is perpendicular to both \(\textbf{A}\) and \(\textbf{B}\).
  • This means if you imagine both vectors \(\textbf{A}\) and \(\textbf{B}\) as arrows in space, \(\textbf{A} \times \textbf{B}\) sticks out at a perpendicular angle to the plane created by \(\textbf{A}\) and \(\textbf{B}\).
Understanding perpendicular vectors is crucial because they are orthogonal, meaning they work independently in space and can define the boundaries of movements in the multi-dimensional space.
Determinant in Vectors
The determinant is a key feature in mathematics, particularly in linear algebra, and it plays a vital role when dealing with vectors. In the context of vectors, it's mostly used to compute the cross product for two vectors.Let's see how determinants help compute the cross product:
  • The cross product \(\textbf{A} \times \textbf{B}\) is obtained by constructing a 3x3 determinant involving unit vectors \(\hat{\textbf{i}}, \hat{\textbf{j}}, \hat{\textbf{k}}\) and the components of vectors \(\textbf{A}\) and \(\textbf{B}\).
  • For \(\textbf{A} = 2 \hat{\textbf{i}} + 3 \hat{\textbf{j}} + 6 \hat{\textbf{k}}\) and \(\textbf{B} = 3 \hat{\textbf{i}} - 6 \hat{\textbf{j}} + 2 \hat{\textbf{k}}\), the determinant is:\[\mathbf{A} \times \mathbf{B} = \begin{vmatrix}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \ 2 & 3 & 6 \ 3 & -6 & 2 \end{vmatrix}\]
  • This expands to generate a vector: \[\hat{\mathbf{i}}(3 \cdot 2 - 6 \cdot (-6)) - \hat{\mathbf{j}}(2 \cdot 2 - 6 \cdot 3) + \hat{\mathbf{k}}(2 \cdot (-6) - 3 \cdot 3)\]
  • Simplifying, you find: \[42 \hat{\mathbf{i}} + 14 \hat{\mathbf{j}} - 21 \hat{\mathbf{k}}\]}.
Using determinants in vectors can look complex, but breaking it down step-by-step gives you the cross product, which finds the vector that is perpendicular to both initial vectors.

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Most popular questions from this chapter

Three vectors \(\mathbf{A}, \mathbf{B}\) and \(\mathbf{C}\) add up to zero. Find which is false (a) \((A \times B) \times C\) is not zero unless \(B, C\) are parallel (b) \((A \times B) \cdot C\) is not zero unless \(B, C\) are parallel (c) If \(A, B, C\) defined a plane, \((A \times B) \times C\) is in that plane (d) \((\mathrm{A} \times \mathrm{B}) \cdot \mathrm{C}=|\mathrm{A}||\mathrm{B}||\mathrm{C}| \rightarrow C^{2}=A^{2}+B^{2}\)

Given \(\mathbf{R}=\mathbf{A}+\mathbf{B}\) and \(R=A=B .\) The angle between \(\mathbf{A}\) and \(\mathbf{B}\) is (a) \(60^{\circ}\) (b) \(90^{\circ}\) (c) \(120^{\circ}\) (d) \(180^{\circ}\)

The resultant of two vectors \(\mathbf{A}\) and \(\mathbf{B}\) is perpendicular to the vector \(\mathbf{A}\) and its magnitude is equal to half of the magnitude of vector \(\mathbf{B}\). Then, the angle between \(\mathbf{A}\) and \(\mathbf{B}\) is (a) \(30^{\circ}\) (b) \(45^{\circ}\) (c) \(150^{\circ}\) (d) \(120^{\circ}\)

The resultant of two forces, each \(\mathbf{P}\), acting at an angle \(\theta\) is (a) \(2 P \sin \frac{\theta}{2}\) (b) \(2 P \cos \frac{\theta}{2}\) (c) \(2 P \cos \theta\) (d) \(P \sqrt{2}\)

The torque of a force \(\mathbf{F}=-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) acting at a point is \(\tau .\) If the position vector of the point is \(7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\), then \(\tau\) is (a) \(7 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}\) (b) \(14 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) (c) \(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}\) (d) \(14 \hat{\mathrm{i}}-38 \hat{\mathrm{j}}+16 \hat{\mathrm{k}}\)
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