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Tutorial 3: Vector Algebra I

  1. Sketch the two points P(3,−1,5)P(3,-1,5) and Q(2,1,−1)Q(2,1,-1) in three-dimensional space and find the distance between the two points.

    Solution
    Figure for Answer 1

    Distance between the two points:

    ∣PQ→∣=(x2−x1)2+(y2−y1)2+(z2−z1)2=(2−3)2+(1−(−1))2+(−1−5)2=41\begin{aligned} |\overrightarrow{P Q}|&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\\ &=\sqrt{(2-3)^2+(1-(-1))^2+(-1-5)^2}\\ &=\sqrt{41} \end{aligned}

  2. For the position vector a∼=⟨2,4⟩\underset{\sim}{a}=\langle 2,4\rangle, compute 3a∼,12a∼3 \underset{\sim}{a}, \frac{1}{2} \underset{\sim}{a}, and −2∼a-\underset{\sim}{2} a. Sketch all four vectors on the same axis system. Discuss the effects of scalar multiplication on the magnitude and direction of the original vector.

    Solution
    Figure for Answer 2
    • The scalar multiplication affects the magnitude of vectors if the scalar is a negative value. For example, −2a∼-2\underset{\sim }{a} switch the direction in exactly the opposite direction.
    • The scalar multiplication affects the magnitude of vectors if the scalar is not equal to 1.
    • When scalar >1> 1, the length of the vector is stretched. (magnitude increased)
    • When scalar <1< 1, the length of the vector is shrinked. (magnitude decteased)

  3. Two vectors are given as OP→=i∼+3j∼−7k∼\overrightarrow{O P}=\underset{\sim}{i}+3 \underset{\sim}{j}-7 \underset{\sim}{k} and OQ→=5∼i−2j∼+4k∼\overrightarrow{O Q}=\underset{\sim}{5} i-2 \underset{\sim}{j}+4 \underset{\sim}{k}

    i. Find the unit vector in the direction of PQ→\overrightarrow{P Q}

    Solution
    PQ→=OQ→−OP→=⟨5,−2,4⟩−⟨1,3,−7⟩=⟨4,−5,11⟩\begin{aligned} \overrightarrow{PQ} & =\overrightarrow{OQ} -\overrightarrow{OP}\\ & =\langle 5,-2,4\rangle -\langle 1,3,-7\rangle \\ & =\langle 4,-5,11\rangle \end{aligned}u^PQ=PQ→∣PQ→∣=⟨4,−5,11⟩(4)2+(−5)2+(11)2=192⟨4,−5,11⟩\begin{aligned} \hat{u}_{PQ} & =\frac{\overrightarrow{PQ}}{| \overrightarrow{PQ}| }\\ & =\frac{\langle 4,-5,11\rangle }{\sqrt{( 4)^{2} +( -5)^{2} +( 11)^{2}}}\\ & =\frac{1}{9\sqrt{2}} \langle 4,-5,11\rangle \end{aligned}

    ii. Find the direction cosines of PQ→\overrightarrow{P Q}

    Solution

    Comparing each component, we obtain:

    • For component i∼\underset{\sim }{i}, cos⁡α=492\cos \alpha =\frac{4}{9\sqrt{2}}
    • For component j∼\underset{\sim }{j}, cos⁡β=−592\cos \beta =\frac{-5}{9\sqrt{2}}
    • For component k∼\underset{\sim }{k}, cos⁡γ=1192\cos \gamma =\frac{11}{9\sqrt{2}}

    iii. Find the vector with magnitude of 5 in the direction of QP→\overrightarrow{Q P} in polar form

    Solution
    cos⁡α=492⟹ α=71.68∘ cos⁡β=−592⟹ β=113.13∘cos⁡γ=1192⟹ γ=30.20∘\begin{aligned} \cos \alpha =\frac{4}{9\sqrt{2}} & \Longrightarrow \ \alpha =71.68^\circ \\ \ \cos \beta =\frac{-5}{9\sqrt{2}} & \Longrightarrow \ \beta =113.13^\circ \\ \cos \gamma =\frac{11}{9\sqrt{2}} & \Longrightarrow \ \gamma =30.20^\circ \end{aligned}

    Direction PQ→\overrightarrow{PQ} is in the direction of its unit vector, ⟨cos⁡(71.68∘),cos⁡(113.13∘),cos⁡(30.20∘)⟩\langle \cos( 71.68^\circ ) ,\cos( 113.13^\circ ) ,\cos( 30.20^\circ ) \rangle.

    Direction QP→\overrightarrow{QP} is in the direction of its unit vector,

    ⟨cos⁡(180−71.68∘),cos⁡(180−113.13∘),cos⁡(180−30.20∘)⟩=⟨cos⁡(108.32∘),cos⁡(66.87∘),cos⁡(149.80∘)⟩\begin{aligned} &\langle \cos( 180-71.68^\circ ) ,\cos( 180-113.13^\circ ) ,\cos( 180-30.20^\circ ) \rangle \\ &=\langle \cos( 108.32^\circ ) ,\cos( 66.87^\circ ) ,\cos( 149.80^\circ ) \rangle \end{aligned}

    Vector of magintude 5 in the direction QP→\overrightarrow{QP} is 5⟨cos⁡(108.32∘),cos⁡(66.87∘),cos⁡(149.80∘)⟩5\langle \cos( 108.32^\circ ) ,\cos( 66.87^\circ ) ,\cos( 149.80^\circ ) \rangle.

  4. Two points are given as A(1,2)A(1,2) and B(3,4)B(3,4).

    i. Find the vector equation of line LL that is passing through point AA and BB.

    Solution

    Direction of vector is parallel with AB→=OB→−OA→=⟨3,4⟩−⟨1,2⟩=⟨2,2⟩\overrightarrow{AB} =\overrightarrow{OB} -\overrightarrow{OA} =\langle 3,4\rangle -\langle 1,2\rangle =\langle 2,2\rangle

    Vector equation of line LL is r∼=a∼+tv∼=⟨1,2⟩+t⟨2,2⟩\underset{\sim }{r} =\underset{\sim }{a} +t\underset{\sim }{v} =\langle 1,2\rangle +t\langle 2,2\rangle

    ii. Sketch the line for t=0:1:5t=0: 1: 5 and indicate its direction and initial point.

    Solution
    ttr∼\underset{\sim }{r}
    0⟨1,2⟩\langle 1,2\rangle
    1⟨3,4⟩\langle 3,4\rangle
    2⟨5,6⟩\langle 5,6\rangle
    3⟨7,8⟩\langle 7,8\rangle
    4⟨9,10⟩\langle 9,10\rangle
    5⟨11,12⟩\langle 11,12\rangle
    Figure for Answer 4b

  5. If a unit vector a⃗\vec{a} makes an angle of π/3\pi / 3 with i,π/4i, \pi / 4 with jj and acute angle θ\theta with kk, find θ\theta and the components of a⃗\vec{a}.

    Solution

    Let α,β,γ\alpha ,\beta ,\gamma to represent angles of a⃗\vec{a}, and a⃗=a1i+a2j+a3k\vec{a} =a_{1} i+a_{2} j+a_{3} k.

    cos⁡α=a1∣a⃗∣cos⁡(π3)=a11a1=12\begin{aligned} \cos \alpha & =\frac{a_{1}}{| \vec{a}| }\\ \cos\left(\frac{\pi }{3}\right) & =\frac{a_{1}}{1}\\ a_{1} & =\frac{1}{2} \end{aligned}cos⁡β=a2∣a⃗∣cos⁡(π4)=a21a2=12\begin{aligned} \cos \beta & =\frac{a_{2}}{| \vec{a}| }\\ \cos\left(\frac{\pi }{4}\right) & =\frac{a_{2}}{1}\\ a_{2} & =\frac{1}{\sqrt{2}} \end{aligned}cos⁡γ=a3∣a⃗∣cos⁡θ=a31a3=cos⁡θ\begin{aligned} \cos \gamma & =\frac{a_{3}}{| \vec{a}| }\\ \cos \theta & =\frac{a_{3}}{1}\\ a_{3} & =\cos \theta \end{aligned}

    ∴ a⃗=12i+12j+cos⁡θk\therefore \ \vec{a} =\frac{1}{2} i+\frac{1}{\sqrt{2}} j+\cos \theta k

    ∣a⃗∣=(12)2+(12)2+(cos⁡θ)21=14+12+cos⁡2θ1=34+cos⁡2θcos⁡2θ=14cos⁡θ=±12\begin{aligned} | \vec{a}| & =\sqrt{\left(\frac{1}{2}\right)^{2} +\left(\frac{1}{\sqrt{2}}\right)^{2} +(\cos \theta )^{2}}\\ 1 & =\sqrt{\frac{1}{4} +\frac{1}{2} +\cos^{2} \theta }\\ 1 & =\frac{3}{4} +\cos^{2} \theta \\ \cos^{2} \theta & =\frac{1}{4}\\ \cos \theta & =\pm \frac{1}{2} \end{aligned}

    Since θ\theta is given as an acute angle, cos⁡θ>0\cos\theta>0, hence cos⁡θ=12\cos\theta=\dfrac{1}{2} and θ=60∘\theta=60^\circ.

    ∴ θ=60∘,a⃗=12i+12j+12k\therefore \ \theta=60^\circ,\qquad \vec{a} =\frac{1}{2} i+\frac{1}{\sqrt{2}} j+\frac{1}{2} k

  6. If a⃗\vec{a} is a unit vector and (x⃗−a⃗)⋅(x⃗+a⃗)=8(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8, find ∣x⃗∣|\vec{x}|.

    Solution

    Since a⃗\vec{a} is a unit vector, ∣a⃗∣=1\vert \vec{a}\vert =1.

    (x⃗−a⃗)⋅(x⃗+a⃗)=8x⃗⋅x⃗+x⃗⋅a⃗−a⃗⋅x⃗−a⃗⋅a⃗=8∣x⃗∣2−∣a⃗∣2=8∣x⃗∣2−1=8∣x⃗∣2=9∣x⃗∣=3\begin{aligned} (\vec{x} -\vec{a}) \cdot (\vec{x} +\vec{a}) & =8\\ \vec{x} \cdot \vec{x} +\vec{x} \cdot \vec{a} -\vec{a} \cdot \vec{x} -\vec{a} \cdot \vec{a} & =8\\ | \vec{x}| ^{2} -| \vec{a}| ^{2} & =8\\ | \vec{x}| ^{2} -1 & =8\\ | \vec{x}| ^{2} & =9\\ | \vec{x}| & =3 \end{aligned}

    Note that magnitude of vector is non-negative.

  7. Find the gradient for f=(2x2+y)/(x2−y2)f=\left(2 x^{2}+y\right) /\left(x^{2}-y^{2}\right).

    Solution
    ddx(f)=(x2−y2)(4x)−(2x2+y)(2x)(x2−y2)2\frac{d}{d x}(f)=\frac{\left(x^2-y^2\right)(4 x)-\left(2 x^2+y\right)(2 x)}{\left(x^2-y^2\right)^2}ddx(f)=2xy(−2y−1)(x2−y2)2\frac{d}{d x}(f)=\frac{2 x y(-2 y-1)}{\left(x^2-y^2\right)^2}

    and

    ddy(f)=(x2−y2)(1)−(2x2+y)(−2y)(x2−y2)2\frac{d}{d y}(f)=\frac{\left(x^2-y^2\right)(1)-\left(2 x^2+y\right)(-2 y)}{\left(x^2-y^2\right)^2}ddy(f)=x2+4x2y+y2(x2−y2)2\frac{d}{d y}(f)=\frac{x^2+4 x^2 y+y^2}{\left(x^2-y^2\right)^2}

    Hence

    ∇f=2xy(−2y−1)(x2−y2)2i+x2+4x2y+y2(x2−y2)2j\nabla f=\frac{2 x y(-2 y-1)}{\left(x^2-y^2\right)^2} \boldsymbol{i}+\frac{x^2+4 x^2 y+y^2}{\left(x^2-y^2\right)^2} \boldsymbol{j}

  8. Calculate the divergence of the following vector fields of F(x,y)F(x, y) and G(x,y)G(x, y);

    a. F=y3i+xyjF=y^{3} \boldsymbol{i}+x y \boldsymbol{j}

    Solution
    ∇⋅F(x,y)=∂F1∂x+∂F2∂y=∂∂xy3+∂∂yxy=0+x=x\begin{aligned} \nabla \cdot \boldsymbol{F}(x, y) &=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y} \\ &=\frac{\partial}{\partial x} y^3+\frac{\partial}{\partial y} x y=0+x=x \end{aligned}

    b. G=4yx2i+sin⁡(y)j+3kG=\frac{4 y}{x^{2}} \boldsymbol{i}+\sin (y) \boldsymbol{j}+3 \boldsymbol{k}

    Solution
    ∇⋅G=∂G1∂x+∂G2∂y+∂G3∂z=∂∂x(4yx2)+∂∂ysin⁡(y)+∂∂z3=4y×∂∂xx−2+cos⁡(y)=4y×(−2)x−2−1+cos⁡(y)=−8yx−3+cos⁡(y)=−8yx3+cos⁡(y)\begin{aligned} \nabla \cdot \boldsymbol{G} &=\frac{\partial G_1}{\partial x}+\frac{\partial G_2}{\partial y}+\frac{\partial G_3}{\partial z} \\ &=\frac{\partial}{\partial x}\left(\frac{4 y}{x^2}\right)+\frac{\partial}{\partial y} \sin (y)+\frac{\partial}{\partial z} 3 \\ &=4 y \times \frac{\partial}{\partial x} x^{-2}+\cos (y)=4 y \times(-2) x^{-2-1}+\cos (y) \\ &=-8 y x^{-3}+\cos (y)=-\frac{8 y}{x^3}+\cos (y) \end{aligned}

    c. G=exi+ln⁡(xy)j+exyzkG=e^{x} \boldsymbol{i}+\ln (x y) \boldsymbol{j}+e^{x y z} \boldsymbol{k}

    Solution
    ∇⋅G=∂G1∂x+∂G2∂y+∂G3∂z=∂∂xex+∂∂yln⁡(xy)+∂∂zexyz=ex+∂∂y(ln⁡(x)+ln⁡(y))+exyz×∂∂z(xyz)=ex+1y+xyexyz\begin{aligned} \nabla \cdot \boldsymbol{G} &=\frac{\partial G_1}{\partial x}+\frac{\partial G_2}{\partial y}+\frac{\partial G_3}{\partial z} \\ &=\frac{\partial}{\partial x} \mathrm{e}^x+\frac{\partial}{\partial y} \ln (x y)+\frac{\partial}{\partial z} \mathrm{e}^{x y z} \\ &=\mathrm{e}^x+\frac{\partial}{\partial y}(\ln (x)+\ln (y))+\mathrm{e}^{x y z} \times \frac{\partial}{\partial z}(x y z) \\ &=\mathrm{e}^x+\frac{1}{y}+x y \mathrm{e}^{x y z} \end{aligned}

  9. Calculate the curl of the following vector fields of F(x,y,z)F(x, y, z);

    a. F=3x2i+2zj−xkF=3 x^{2} \boldsymbol{i}+2 z \boldsymbol{j}-x \boldsymbol{k}

    Solution
    ∇×F=∣ijk∂∂x∂∂y∂∂z3x22z−x∣=(∂(−x)∂y−∂(2z)∂z)i−(∂(−x)∂x−∂(3x2)∂z)j+(∂(2z)∂x−∂(3x2)∂y)k=(0−2)i−(−1−0)j+(0−0)k=−2i+j\begin{aligned} \nabla \times \boldsymbol{F} &=\left|\begin{array}{ccc}i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 3 x^2 & 2 z & -x\end{array}\right| \\ &=\left(\frac{\partial(-x)}{\partial y}-\frac{\partial(2 z)}{\partial z}\right) i-\left(\frac{\partial(-x)}{\partial x}-\frac{\partial\left(3 x^2\right)}{\partial z}\right) j +\left(\frac{\partial(2 z)}{\partial x}-\frac{\partial\left(3 x^2\right)}{\partial y}\right) k \\ &=(0-2) i-(-1-0) j+(0-0) k \\ &=-2 i+j \end{aligned}

    b. F=y3i+xyj−zkF=y^{3} \boldsymbol{i}+x y \boldsymbol{j}-z \boldsymbol{k}

    Solution
    ∇×F=(∂F3∂y−∂F2∂z)i−(∂F3∂x−∂F1∂z)j+(∂F2∂x−∂F1∂y)k=(∂(−z)∂y−∂(xy)∂z)i−(∂(−z)∂x−∂(y3)∂z)j+(∂(xy)∂x−∂(y3)∂y)k=0i−0j+(y−3y2)k=(y−3y2)k\begin{aligned} \nabla \times \boldsymbol{F}=&\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right) i-\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right) j+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right) \boldsymbol{k} \\ =&\left(\frac{\partial(-z)}{\partial y}-\frac{\partial(x y)}{\partial z}\right) i-\left(\frac{\partial(-z)}{\partial x}-\frac{\partial\left(y^3\right)}{\partial z}\right) j +\left(\frac{\partial(x y)}{\partial x}-\frac{\partial\left(y^3\right)}{\partial y}\right) \boldsymbol{k} \\ =& 0 i-0 j+\left(y-3 y^2\right) \boldsymbol{k}\\ =&\left(y-3 y^2\right) \boldsymbol{k} \end{aligned}

    i.e., the curl vector is in the kk direction.

    с. F=(1+y+z2)i+(exyz)j−(xyz)kF=\left(1+y+z^{2}\right) \boldsymbol{i}+\left(e^{x y z}\right) \boldsymbol{j}-(x y z) \boldsymbol{k}

    Solution
    curl⁡(F)=∇×F=(∂∂y(−xyz)−∂∂z(exyz))i⃗+(∂∂z(1+y+z2)−∂∂x(−xyz))j⃗+(∂∂x(exyz)−∂∂y(1+y+z2))k⃗\operatorname{curl}(\mathbf{F})=\nabla \times \mathbf{F}=\left(\frac{\partial}{\partial y}(-x y z)-\frac{\partial}{\partial z}\left(e^{x y z}\right)\right) \vec{i}+\left(\frac{\partial}{\partial z}\left(1+y+z^2\right)-\frac{\partial}{\partial x}(-x y z)\right) \vec{j}+\left(\frac{\partial}{\partial x}\left(e^{x y z}\right)-\frac{\partial}{\partial y}\left(1+y+z^2\right)\right) \vec{k}curl⁡(F)=−(xz+xyexyz)i⃗+(2z+yz)j⃗+(yzexyz−1)k⃗\operatorname{curl}(\mathbf{F})=-\left(x z+x y e^{x y z}\right) \vec{i}+(2 z+y z) \vec{j}+\left(y z e^{x y z}-1\right) \vec{k}