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Chapter 3: Q79P (page 61)

In Fig. 3-38 , the magnitude of $\overset{⇀}{a} is 4.3$ , the magnitude of $\overset{⇀}{b} is 5.4$ , and $ϕ = 46 °$ . Find the area of the triangle contained between the two vectors and the thin diagonal line.

Short Answer

Expert verified

Area of a triangle is8.4

Step by step solution

01

To understand the concept

The magnitude of the vector product is the area of parallelogram spanned by these two vectors. Therefore, vector product is used to compute the area of a triangle.

Here the area of a triangle is given by,

$A = \frac{1}{2} |\overset{⇀}{a} \times \overset{⇀}{b}| = \frac{1}{2} ab sinϕ (1) Given, \overset{⇀}{a} = 4.3 \overset{⇀}{b} = 5.4 The angle between the vector \overset{⇀}{a} and \overset{⇀}{b} is ϕ = 46 °$

02

To find the area of a triangle

Substituting the values of $\overset{⇀}{a}, \overset{⇀}{b}$ , and $ϕ$ in equation (i), the area of a triangle is given by

$A = \frac{1}{2} absinϕ = \frac{1}{2} (4.3) (5.4) \sin (46)$

Thus area of triangle $A = 8.4$

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Most popular questions from this chapter

In Fig. $3 - 27$ , a heavy piece of machinery is raised by sliding it a distance $d = 12.5 m$ along a plank oriented at angle $θ = 20.0 °$ to the horizontal. How far is it moved (a) vertically and (b) horizontally?

Here are three displacements, each measured in meters: ${\vec{d}}_{1} = 4.0 \overset{⏜}{i} + 5.0 \overset{⏜}{j} - 6.0 \overset{⏜}{k}, {\vec{d}}_{2} = - 1.0 \overset{⏜}{i} + 2.0 \overset{⏜}{j} + 3.0 \overset{⏜}{k}$ ,and ${\vec{d}}_{3} = 4.0 \overset{⏜}{i} + 3.0 \overset{⏜}{j} + 2.0 \overset{⏜}{k}$ . (a) What is $\vec{r} = {\vec{d}}_{1} + {\vec{d}}_{2} + {\vec{d}}_{3}$ ? (b) What is the angle between $\vec{r}$ and the positive z axis? (c) What is the component of $d_{1}$ along the direction of ${\vec{d}}_{2}$ (d) What is the component of ${\vec{d}}_{1}$ that is perpendicular to the direction of $\vec{d_{2}}$ and in the plane of role="math" localid="1658465314757" ${\vec{d}}_{1}$ and ${\vec{d}}_{2}$ (Hint: For (c), consider Eq 3-20.and Fig, 3-18; for (d), consider Eq.3-24.)

Here are three vectors in meters:

$\vec{d_{1}} = - 3.0 \hat{i} + 3.0 \hat{j} + 2.0 \hat{k}$
localid="1654741448647" $\vec{d_{2}} = - 2.0 \hat{i} + 4.0 \hat{j} + 2.0 \hat{k}$

localid="1654741501014" $\vec{d_{3}} = 2.0 \hat{i} + 3.0 \hat{j} + 1.0 \hat{k}$

What result from (a)localid="1654740492420" $^{\vec{d_{1}} . (\vec{d_{2}} + \vec{d_{3}})}$ , (b) $^{\vec{d_{1}} . (\vec{d_{2}} \times \vec{d_{3}})}$ , and (c) localid="1654740727403" $^{d_{1} \times (d_{2} + d_{3})}$ ?

Vector $\overset{⇀}{A}$ which is directed along an x axis, which to be added to

vector $\overset{⇀}{B}$ , which has a magnitude of 7.0 m.The sum is a third vector that

is directed along the y axis, with a magnitude that is 3.0 times that of $\overset{⇀}{A}$ . What is that magnitude of $\overset{⇀}{A}$ ?

Let $\hat{i}$ be directed to the east, $\hat{j}$ be directed to the north, and be directed upward.What are the values of products (a) $\hat{i} . \hat{k}, (b) (- \hat{k}) . (\hat{j})$ , and (c) $(\hat{j}) . (- \hat{j})$ ? What are the directions (such as east or down) of products (d) $\hat{k} \times \hat{j}, (e) (- \hat{i}) \times (- \hat{j}), and (f) (- \hat{k}) \times (- \hat{j})$ ?

See all solutions

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