Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q16E

Expert-verified
Essential Calculus: Early Transcendentals
Found in: Page 564
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

(a) Find the magnitude of cross product\(|a \times b|\).

(b) Check whether the components of \(a \times b\) are positive, negative or 0.

(a) The value of the cross product \(|a \times b|\) is 6.

(b) The \(x\) component of \({\rm{a}} \times {\rm{b}}\) is a positive, \(y\) component of \({\rm{a}} \times {\rm{b}}\) is a negative, and \(z\) component of \({\rm{a}} \times {\rm{b}}\) is zero.

See the step by step solution

Step by Step Solution

Step 1: Formula used

If \(\theta \) is the angle between vectors \({\bf{u}}\) and \({\bf{v}}\), then the cross product is,

\(|u \times v| = |u||v|\sin \theta \ldots \ldots \ldots (1)\)

Here,

|u| is the magnitude of vector\({\bf{u}}\) and

|v| is the magnitude of vector\({\bf{v}}\).

Right-hand rule:

The fingers of the right hand curl in the direction of rotation from \(a\) to \(b\), then the thumb points are in the direction of \(a \times b\). The rotation angle should be less than\({180^^\circ }\).

Step 2: Find the \(|a \times b|\)

(a)

As given, \(|{\rm{a}}| = 3,|\;{\rm{b}}| = 2\).

So, the value of angle \((\theta )\) is \({90^^\circ }\)by applying the right hand rule

Substitute 3 for \(|{\rm{a}}|,2\) for \(|{\rm{b}}|\), and \({90^^\circ }\) for \(\theta \) in \(|{\rm{a}} \times {\rm{b}}| = |{\rm{a}}||{\rm{b}}|\sin \theta \)

\(\begin{array}{l}|a \times b| = (3)(2)\sin \left( {{{90}^^\circ }} \right)\\|a \times b| = 6(1)\\|a \times b| = 6\end{array}\)

Thus, the magnitude of cross product \(|a \times b|\) is 6.

Step 3: Check whether the components of \(a \times b\) are positive, negative or 0

(b)

From part (a), it is observed that, as vector a lies on the x y- plane, the \(z\)-component of \(a \times b\)is zero.

When the direction of vector \({\rm{a}} \times {\rm{b}}\) is opposite to the direction of \(y\) axis, the \(y\)-component of \({\rm{a}} \times {\rm{b}}\) is negative.

As the direction of vector \({\rm{a}} \times {\rm{b}}\) is in the direction of \(x\) axis, the \(x\)-component of \({\rm{a}} \times {\rm{b}}\) is positive.

Therefore, the \(x\) component of \({\rm{a}} \times {\rm{b}}\) is a positive, \(y\) component of \({\rm{a}} \times {\rm{b}}\) is a negative, and \(z\) component of \(a \times b\) is zero.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.