šŸ“ Unit 7: Vector Applications

Dot Product Ā· Cross Product Ā· Forces Ā· Work Ā· Projections
šŸ”„ Not Graded
Purpose: Self-check on dot/cross products and physical applications.
Score: 0 / 12
Topic 7.1 ā€” Dot Product
Question 1
For \(\vec u=(2,3,-1)\) and \(\vec v=(4,-2,5)\), find \(\vec u\cdot\vec v\).
Solution:
\(8-6-5=-3\).
Question 2
Vectors \((1,2)\) and \((4,-2)\) are:
Solution:
Dot \(=4-4=0\) ā†’ perpendicular.
Question 3
Angle between \(\vec u=(1,2,2)\) and \(\vec v=(3,0,4)\), in degrees.
Solution:
\(\vec u\cdot\vec v=3+0+8=11\); \(|\vec u|=3,\,|\vec v|=5\); \(\cos\theta=11/15\) ā†’ \(\theta\approx 42.83Ā°\).
Topic 7.2 ā€” Projections
Question 4
Scalar projection of \(\vec u=(3,4)\) onto \(\vec v=(1,0)\).
Solution:
\(\vec u\cdot\vec v/|\vec v|=3/1=3\).
Question 5
Vector projection of \((2,3,-1)\) onto \((1,0,0)\). First component:
Solution:
\(\frac{(2)(1)+0+0}{1}\hat v=(2,0,0)\). First component = 2.
Topic 7.3 ā€” Cross Product
Question 6
For \(\vec u=(1,0,0)\) and \(\vec v=(0,1,0)\), \(\vec u\times\vec v=\)
Solution:
\(\hat i\times\hat j=\hat k=(0,0,1)\).
Question 7
For \(\vec a=(2,1,-3),\,\vec b=(1,4,2)\), the third component of \(\vec a\times\vec b\) is:
Solution:
\((a_1 b_2 - a_2 b_1)=2(4)-1(1)=8-1=7\).
Question 8
The area of the parallelogram with sides \((1,2,0)\) and \((3,1,0)\) is:
Solution:
\(\vec u\times\vec v=(0,0,1\cdot 1-2\cdot 3)=(0,0,-5)\); magnitude = 5.
Topic 7.4 ā€” Work & Force Applications
Question 9
A force \(\vec F=(5,3,-2)\) N moves a body through displacement \(\vec d=(2,1,4)\) m. Work done (J):
Solution:
\(W=\vec F\cdot\vec d=10+3-8=5\) J.
Question 10
A 50 N force at 30Ā° above horizontal pulls a sled 20 m horizontally. Work done:
Solution:
\(W=Fd\cos\theta=50(20)\cos 30Ā°=1000\cdot 0.866\approx 866\) J.
Question 11
Torque magnitude when \(\vec r=(0,2,0)\) m and \(\vec F=(5,0,0)\) N:
Solution:
\(\vec\tau=\vec r\times\vec F=(0,0,2\cdot 0-2\cdot 5)? \) Compute: \(r\times F=(2\cdot 0-0\cdot 0,\,0\cdot 5-0\cdot 0,\,0\cdot 0-2\cdot 5)=(0,0,-10)\); magnitude 10 NĀ·m.
Question 12
If \(\vec u\cdot\vec v=0\) and neither vector is zero, the angle between them is:
Solution:
\(\cos 90Ā°=0\) ā†’ perpendicular.

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