Multivariable Calculus 12.3 Textbook Notes

Status: Adult

Formulas

• The work done by a constant force $F$ acting through a displacement $D=\overline{PQ}$ is $W=F\cdot D$

Find definitions/descriptions for the following terms:

• The dot product between two vectors
  • The dot product $u\cdot v$ (“u dot v”) of vectors $u=<u_1,u_2,u_3>$ and $v=<v_1,v_2,v_3>$ is the scalar $u\cdot v=u_1{v}_1+u_2{v}_2+u_3{v}_3$
• Compute angle between two vectors in terms of their dot product
  • The angle $\theta$ between two nonzero vectors $u=<u_1,u_2,u_3>$ and $v=<v_1,v_2,v_3>$ is given by $\theta =\arccos \left(\frac{u_1{v}_1+u_2{v}_2+u_3{v}_3}{|u||v|}\right)=\arccos \left(\frac{u\cdot v}{|u||v|}\right)$
• Relation of dot product between vectors and the law of cosines

  • Transclude of Parallelogram-Law-of-Addition-of-Vectors

    $|w{|}^2=|u{|}^2+|v{|}^2-2|u||v|\cos \theta$ $2|u||v|\cos \theta =|u{|}^2+|v{|}^2-|w{|}^2$ Because $w=u-v$, the component form of $w$ is $<u_1-v_1,u_2-v_2,u_3-v_3>$ Therefore, $|u{|}^2=(\sqrt{u_1^2+u_2^2+u_3^2}{)}^2=u_1^2+u_2^2+u_3^2$ $|v{|}^2=(\sqrt{v_1^2+v_2^2+v_3^2}{)}^2=v_1^2+v_2^2+v_3^2$ $|w{|}^2=(\sqrt{(u_1-v_1{)}^2+(u_2-v_2{)}^2+(u_3-v_3{)}^2}{)}^2$ $=(u_1-v_1{)}^2+(u_2-v_2{)}^2+(u_3-v_3{)}^2$ $=u_1^2-2u_1{v}_1+v_1^2+u_2^2-2u_2{v}_2+v_2^2+u_3^2-2u_3{v}_3+v_3^2$ And $|u{|}^2+|v{|}^2-|w{|}^2=2(u_1{v}_1+u_2{v}_2+u_3{v}_3)$ Therefore, $2|u||v|\cos \theta =|u{|}^2+|v{|}^2-|w{|}^2=2(u_1{v}_1+u_2{v}_2+u_3{v}_3)$ $|u||v|\cos \theta =u_1{v}_1+u_2{v}_2+u_3{v}_3$ $\cos \theta =\left(\frac{u_1{v}_1+u_2{v}_2+u_3{v}_3}{|u||v|}\right)=\left(\frac{u\cdot v}{|u||v|}\right)$ $\theta =\arccos \left(\frac{u\cdot v}{|u||v|}\right)$
• Orthogonal relation between two vectors in terms of their dot product
  • Vectors $u$ and $v$ are orthogonal if $u\cdot v=0$
• Algebraic properties of the dot product
  • $u\cdot v=v\cdot u$
  • $(cu)\cdot v=u\cdot (cv)=c(u\cdot v)$
  • $u\cdot (v+w)=u\cdot v+u\cdot w$
  • $u\cdot u=|u{|}^2$
  • $0\cdot u=0$
• The scalar component of u in the direction of v and the (orthogonal) projection of u in the direction of v.
  • The vector projection of $u$ onto $v$ is the vector $pro{j}_{v}u=\left(\frac{u\cdot v}{|v{|}^2}\right)=\left(\frac{u\cdot v}{|v|}\right)\frac{v}{|v|}$
  • The scalar component of $u$ in the direction of $v$ is the scalar $|u|\cos \theta =\frac{u\cdot v}{|v|}=u\cdot \frac{v}{|v|}$

Reading Questions/Quizzes

• Is $u\cdot v$ always a nonnegative number for any vectors u and v?
  • Because $u\cdot v=|u||v|\cos \theta$, and $\cos \theta$ can be negative, $u\cdot v$ is not always a nonnegative number
• Is the scalar component of u in the direction of v always a nonnegative number for any vectors u and v?
  • Because the scalar component is $|u|\cos \theta$ and $\cos \theta$ can be negative, it is not always a nonnegative number
• Is the scalar component of u in the direction of v a vector?
  • No, a scalar is not a vector
• Is the projection of u in the direction of v a vector or a scalar?
  • Yes, because it is a vector projection
• Is it true that $(u-pro{j}_{v}u)\cdot v=0$?
  • Yes, because $u-pro{j}_{v}u$ represents the orthogonal vector to $v$ and the dot product of two orthogonal vectors is 0, it is true
• Is it true that |u · v| ≤ |u||v| always holds?
  • Yes (see Cauchy-Schwarz Inequality)
• When does u · v = |u||v| hold? When does u · v = −|u||v| hold?
  • The former holds when the vectors are pointing in the same direction, and the latter holds when the vectors are pointing in equally opposite directions
• Is it possible that u · v = 2|u||v| for non-zero vectors u, v?
  • No, because the maximum value for cosine is 1, which means the max value for $u\cdot v=|u||v|\cos \theta$ is $|u||v|$ at $\theta =0+2\pi n$ when $\cos \theta =1$

References

Multivariable Full Textbook.pdf 12.3.pdf