2.4 The cross product (Page 4/16)

Page 4 / 16

Using expansion along the first row to compute a $3 \times 3$ Determinant

Evaluate the determinant $| \begin{matrix} 2 & 5 & −1 \\ −1 & 1 & 3 \\ −2 & 3 & 4 \end{matrix} | .$

We have

\begin{matrix} | \begin{matrix} 2 & 5 & −1 \\ −1 & 1 & 3 \\ −2 & 3 & 4 \end{matrix} | & = 2 | \begin{matrix} 1 & 3 \\ 3 & 4 \end{matrix} | - 5 | \begin{matrix} −1 & 3 \\ −2 & 4 \end{matrix} | - 1 | \begin{matrix} −1 & 1 \\ −2 & 3 \end{matrix} | \\ = 2 (4 - 9) - 5 (−4 + 6) - 1 (−3 + 2) \\ = 2 (−5) - 5 (2) - 1 (−1) = −10 - 10 + 1 \\ = −19. \end{matrix}

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Evaluate the determinant $| \begin{matrix} 1 & −2 & −1 \\ 3 & 2 & −3 \\ 1 & 5 & 4 \end{matrix} | .$

$40$

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Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.

Rule: cross product calculated by a determinant

Let $u = ⟨ u_{1}, u_{2}, u_{3} ⟩$ and $v = ⟨ v_{1}, v_{2}, v_{3} ⟩$ be vectors. Then the cross product $u \times v$ is given by

u \times v = | \begin{matrix} i & j & k \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{matrix} | = | \begin{matrix} u_{2} & u_{3} \\ v_{2} & v_{3} \end{matrix} | i - | \begin{matrix} u_{1} & u_{3} \\ v_{1} & v_{3} \end{matrix} | j + | \begin{matrix} u_{1} & u_{2} \\ v_{1} & v_{2} \end{matrix} | k .

Using determinant notation to find $p \times q$

Let $p = ⟨ −1, 2, 5 ⟩$ and $q = ⟨ 4, 0, −3 ⟩ .$ Find $p \times q .$

We set up our determinant by putting the standard unit vectors across the first row, the components of $u$ in the second row, and the components of $v$ in the third row. Then, we have

\begin{matrix} p \times q & = | \begin{matrix} i & j & k \\ −1 & 2 & 5 \\ 4 & 0 & −3 \end{matrix} | = | \begin{matrix} 2 & 5 \\ 0 & −3 \end{matrix} | i - | \begin{matrix} −1 & 5 \\ 4 & −3 \end{matrix} | j + | \begin{matrix} −1 & 2 \\ 4 & 0 \end{matrix} | k \\ = (−6 - 0) i - (3 - 20) j + (0 - 8) k \\ = −6 i + 17 j - 8 k . \end{matrix}

Notice that this answer confirms the calculation of the cross product in [link] .

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Use determinant notation to find $a \times b,$ where $a = ⟨ 8, 2, 3 ⟩$ and $b = ⟨ −1, 0, 4 ⟩ .$

$8 i - 35 j + 2 k$

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Using the cross product

The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a parallelepiped . The following examples illustrate these calculations.

Finding a unit vector orthogonal to two given vectors

Let $a = ⟨ 5, 2, −1 ⟩$ and $b = ⟨ 0, −1, 4 ⟩ .$ Find a unit vector orthogonal to both $a$ and $b .$

The cross product $a \times b$ is orthogonal to both vectors $a$ and $b .$ We can calculate it with a determinant:

\begin{matrix} a \times b & = | \begin{matrix} i & j & k \\ 5 & 2 & −1 \\ 0 & −1 & 4 \end{matrix} | = | \begin{matrix} 2 & −1 \\ −1 & 4 \end{matrix} | i - | \begin{matrix} 5 & −1 \\ 0 & 4 \end{matrix} | j + | \begin{matrix} 5 & 2 \\ 0 & −1 \end{matrix} | k \\ = (8 - 1) i - (20 - 0) j + (−5 - 0) k \\ = 7 i - 20 j - 5 k . \end{matrix}

Normalize this vector to find a unit vector in the same direction:

‖ a \times b ‖ = \sqrt{{(7)}^{2} + {(−20)}^{2} + {(−5)}^{2}} = \sqrt{474} .

Thus, $⟨ \frac{7}{\sqrt{474}}, \frac{−20}{\sqrt{474}}, \frac{−5}{\sqrt{474}} ⟩$ is a unit vector orthogonal to $a$ and $b .$

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Find a unit vector orthogonal to both $a$ and $b,$ where $a = ⟨ 4, 0, 3 ⟩$ and $b = ⟨ 1, 1, 4 ⟩ .$

$⟨ \frac{−3}{\sqrt{194}}, \frac{−13}{\sqrt{194}}, \frac{4}{\sqrt{194}} ⟩$

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To use the cross product for calculating areas, we state and prove the following theorem.

Area of a parallelogram

If we locate vectors $u$ and $v$ such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by $‖ u \times v ‖$ ( [link] ).

This figure is a parallelogram. One side is represented with a vector labeled “v.” The second side, the base, has the same initial point as vector v and is labeled “u.” The angle between u and v is theta. Also, a perpendicular line segment is drawn from the terminal point of v to vector u. It is labeled “|v|sin(theta).” — The parallelogram with adjacent sides $u$ and $v$ has base $‖ u ‖$ and height $‖ v ‖ sin θ .$

Proof

We show that the magnitude of the cross product is equal to the base times height of the parallelogram.

\begin{matrix} Area of a parallelogram & = base \times height \\ = ‖ u ‖ (‖ v ‖ sin θ) \\ = ‖ u \times v ‖ \end{matrix}

□

Finding the area of a triangle

Let $P = (1, 0, 0), Q = (0, 1, 0), and R = (0, 0, 1)$ be the vertices of a triangle ( [link] ). Find its area.

This figure is the 3-dimensional coordinate system. It has a triangle drawn in the first octant. The vertices of the triangle are points P(1, 0, 0); Q(0, 1, 0); and R(0, 0, 1). — Finding the area of a triangle by using the cross product.

We have $\vec{P Q} = ⟨ 0 - 1, 1 - 0, 0 - 0 ⟩ = ⟨ −1, 1, 0 ⟩$ and $\vec{P R} = ⟨ 0 - 1, 0 - 0, 1 - 0 ⟩ = ⟨ −1, 0, 1 ⟩ .$ The area of the parallelogram with adjacent sides $\vec{P Q}$ and $\vec{P R}$ is given by $‖ \vec{P Q} \times \vec{P R} ‖ :$

\begin{matrix} \vec{P Q} \times \vec{P R} & = & | \begin{matrix} i & j & k \\ −1 & 1 & 0 \\ −1 & 0 & 1 \end{matrix} | = (1 - 0) i - (−1 - 0) j + (0 - (−1)) k = i + j + k \\ ‖ \vec{P Q} \times \vec{P R} ‖ & = & ‖ ⟨ 1, 1, 1 ⟩ ‖ = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3} . \end{matrix}

The area of $Δ P Q R$ is half the area of the parallelogram, or $\sqrt{3} / 2 .$

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Find the area of the parallelogram $P Q R S$ with vertices $P (1, 1, 0), Q (7, 1, 0), R (9, 4, 2),$ and $S (3, 4, 2) .$

$6 \sqrt{13}$

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Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.

Kate Reply

To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.

Muhammed

What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?

Kate Reply

what is the change in momentum of a body?

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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.

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A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate

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8m/s²

Aishat

What is Thermodynamics

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velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.

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A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat

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50 m/s due south east

Someone

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I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.

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Someone

temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....

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about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle

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Another formula for Acceleration

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a=v/t. a=f/m a

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Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.

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like charges repel while unlike charges atttact

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What is specific heat capacity

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Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).

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specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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2.4 The cross product (Page 4/16)

Using expansion along the first row to compute a 3 × 3 Determinant

Rule: cross product calculated by a determinant

Using determinant notation to find p × q

Using the cross product

Finding a unit vector orthogonal to two given vectors

Area of a parallelogram

Proof

Finding the area of a triangle

Questions & Answers

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Using expansion along the first row to compute a $3 \times 3$ Determinant

Using determinant notation to find $p \times q$