Solutions to Questions on Scalar and Cross Products for 3D Vectors
Detailed solutions to questions on Scalar and Cross Products of 3D Vectors are presented.
1) Calculate \( \vec{u} \cdot (\vec{u} \times \vec{v}) \) given that \( \vec{u} = \lt a,b,c \gt \) and \( \vec{v} = \lt d,e,f \gt \).
Solution
The cross product \( \vec{u} \times \vec{v}\) is a vector perpendicular to both vectors \( \vec{u} \) and \( \vec{v} \). Hence the dot product of two perpendicular vectors \( \vec{u} \) and \( \vec{u} \times \vec{v} \) is equal to 0.
\( \vec{u} \cdot (\vec{u} \times \vec{v}) = 0 \)
2) Find \( k \) so that vectors \( \vec{u} = \lt -2,-k,1 \gt \) and \( \vec{v} = <8,-2,-3> \) are perpendicular
Solution
If two vectors \( \vec{u} \) and \( \vec{v}\) are perpendicular then their dot product is equal to 0. Hence
\( \vec{u} \cdot \vec{v} = \lt -2,-k,1 \gt \cdot <8,-2,-3> = 0 \)
Expand the dot product to obtain the equation.
\( (-2)(8)+(-k)(-2)+(1)(-3) = 0 \)
Simplify and solve for \( k \).
\( -16 + 2 k - 3 = 0 \)
\( k = 19/2 \)
3)Find \( k \) so that the vectors \( \vec{u} = \lt -3,2,-2 \gt \), \( \vec{v} = <2,1,k> \) and \( \vec{w} = <-1,3,-5> \) are on the same plane (or coplanar)?
Solution
Any two vectors are on the same plane (or coplanar). If a third vector is on this plane, the volume of the parallelepiped (see formula in Scalar and Cross Products of 3D Vectors ) formed by the 3 vectors is equal to 0. Hence the condition for any 3 (non zero) vectors to be coplanar is
\( \vec{u} \cdot (\vec{v} \times \vec{w} ) = 0 \)
\( \vec{u} \cdot (\vec{v} \times \vec{w} ) \) is called the scalar triple product and is given by.
\( \vec{u} \cdot (\vec{v} \times \vec{w} ) = det \begin{bmatrix} u_x & u_y & u_z \\ v_x& v_y & v_z \\ w_x& w_y & w_z \end{bmatrix} \)
where \( u_x, u_y, u_z, v_x... \) are the components of the three vectors.
We now substitute the components and calculate the determinant.
\( \vec{u} \cdot (\vec{v} \times \vec{w} ) = det \begin{bmatrix} -3 & 2 & -2 \\ 2 & 1 & k \\ -1 & 3 & -5 \end{bmatrix} \) = -3(-5 - 3k) - 2(-10 +k) -2(6 + 1) = 21+7k
For the three vectors to be on the same plane, the determinant found above must be zero. Hence.
\( 21+7k = 0 \)
Solve the above equation for \( k \) to obtain.
\( k = -3 \)
Below are shown the three vectors on the same plane.
4) Find angle \( \theta \) between the vectors \( \vec{u} = \lt 2,0,1 \gt \) and \( \vec{v} = <8,-2,-3> \).
Solution
Use the definition of the scalar product of two vectors \( \vec{u} \) and \( \vec{v} \) and its expression using the vector components.
Definition: \( \vec{u} \cdot \vec{v} = ||\vec{u}|| || \vec{v} || \cos \theta \) , θ is the angle between the two vectors.
The scalar product is also given by.
\( \vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z \) , where \( u_x, v_x, u_y, ...\) are the components of the vectors \( \vec{u} \) and \( \vec{v} \).
Let us calculate the magnitudes \( ||\vec{u}||\) and \( || \vec{v} || \) .
\( ||\vec{u}|| = \sqrt{2^2+0^2+1^2} = \sqrt{5} \)
\( ||\vec{v}|| = \sqrt{8^2+(-2)^2+(-3)^2} = \sqrt{77} \)
We now use the components to find the scalar product.
\( \vec{u} \cdot \vec{v} = (2)(8)+(0)(-2)+(1)(-3) = 13 \)
We now use the definition to find angle θ.
\( \vec{u} \cdot \vec{v} = ||\vec{u}|| || \vec{v} || \cos \theta \)
\( \cos \theta = \dfrac{\vec{u} \cdot \vec{v}}{||\vec{u}|| || \vec{v} ||} = \dfrac{13}{\sqrt{5}\sqrt{77}} \)
\( \theta = \arccos \dfrac{13}{\sqrt{5}\sqrt{77}} = 48.5^{\circ} \)
5) Find the vector projection of \( \vec{u} = \lt -1,-1,1 \gt \) onto \( \vec{v} = <2,1,1> \).
Solution
The vector projection of of \( \vec{u}\) on \( \vec{v}\) is given by (see formula in Scalar and Cross Products of 3D Vectors ):
\( \text{proj}_{\vec{v}}\vec{u} = \dfrac{\vec{u}\cdot\vec{v}}{||v||^2} \vec{v} \)
\( = \dfrac{ \lt -1,-1,1 \gt \cdot <2,1,1>}{2^2+1^2+1^2} <2,1,1> = \dfrac{(-1)(2)+(-1)(1)+(1)(1)}{6} <2,1,1>\)
\( = -\dfrac{2}{6} <2,1,1> = <-2/3,-1/3,-1/3> \)
Vectors \( \vec{u}\) , \( \vec{v}\) and \( \text{proj}_{\vec{v}}\vec{u}\) are shown below.
6) Find that \( k \) so that the points \( A(-1,2,k) \), \( B(-3,6,3) \) and \( C(1,3,6) \) are the vertices of a right triangle with a right angle at \( A \).
Solution
For triangle ABC to be right at A, the vectors \( \vec{AB} \) and \( \vec{AC}\) has to be perpendicular and therefore their scalar product is equal to 0. We start by calculating the components of the vectors.
\( \vec{AB} = <-2,4,3-k> \)
\( \vec{AC} = <2,1,6-k> \)
Scalar product of \( \vec{AB} \) and \( \vec{AC}\) has to be zero.
\( <-2,4,3-k> \cdot <2,1,6-k> = 0 \)
\( -4 + 4 + (3-k)(6-k) = 0 \)
Simplify and solve for k.
Two solutions: \( k = 3 \) and \( k = 6 \)
7) Given vector \( \vec{v} = \lt 3,-1,-2 \gt \), find vector \( \vec{u} \) such that \( \vec{v} \times \vec{u} = <4,2,5> \) and \( ||\vec{u}|| = 3\)
Solution
Let a, b and c be the components of vector \( \vec{u} \). Hence
\( \vec{v} \times \vec{u} = {\begin{vmatrix}\vec{i}& \vec{j} &\vec{k} \\ 3 & -1 & -2 \\ a & b & c \end{vmatrix}} \)
\( = {\begin{vmatrix} -1 & -2 \\ b & c \end{vmatrix}} \vec{i} - {\begin{vmatrix}3 & -2\\ a & c\end{vmatrix}} \vec{j} + {\begin{vmatrix}3 & -1\\ a& b\end{vmatrix}} \vec{k} = (-c+2b)\vec{i} + (-2a-3c)\vec{j} + (3b+a)\vec{k} \)
We now write that the components of \( \vec{v} \times \vec{u} \) and \( <4,2,5> \) are equal as given above. Hence
\( -c+2b = 4 \)
\( -2a-3c = 2 \)
\( 3b+a = 5 \)
Note that the equations in the above system are not independent ( add - 3 times the first equation -c+2b = 4 and the second equation -2a-3c = 2 and you will obtain the an equation equivalent to the third equation 3b+a = 5) and therefore it has many solutions. Let a = t and use second equation to find c.
-2t - 3c = 2
c = (2 + 2t) / (-3)
Let a = t again and use third equation to find b.
3b + t = 5
b = (5 - t) / 3
We now use the condition \( ||\vec{u}|| = 3\) to write the equation.
\( \sqrt{a^2+b^2+c^2} = 3 \)
Square both side of the above equation and substitute a, b and c by their expressions in terms of t.
\( t^2+((5 - t) / 3)^2+((2 + 2t) / (-3))^2 = 9 \)
Multiply all terms of the equation by 9 and simplify.
\( 9 t^2+(5 - t)^2+(2 + 2t)^2 = 81 \)
Expand the above to obtain a quadratic equation and solve it to find
t = 2 and t = -13/7
Hence two solutions for vector \( \vec{u} \).
for t = 2 : \( \vec{u_1} = < t , (5 - t)/3 , (2 + 2t)/(-3) > = <2 , 1, -2> \)
for t = -13/7 : \( \vec{u_2} = < t , (5 - t)/3 , (2 + 2t)/(-3) > = <-13/7 , 16/7 , 4/7> \)
8) Points A, B, C and D forms a parallelogram.
a)Find the coordinates of point D.
b)Find the area of a parallelogram.
Solution
a) Let a, b and c be the coordinates of point D and determine the components of vectors \( \vec{AB} \) and \( \vec{DC} \).
\( \vec{AB} = <2 - 4 , 2 - 6 , 4 - 2 > = <-2 ,-4 ,2> \)
\( \vec{DC} = <-2 - a , -3 - b , 1 - c> \)
For A, B, C and D to form a parallelogram, vectors \( \vec{AB} \) and \( \vec{DC} \) have to be equal (equivalent). Hence the vector equation
\( <-2 , -4 , 2 > = <-2 - a , -3 - b , 1 - c> \)
Which gives 3 algebraic equations.
-2 - a = -2
-3 - b = -4
1 - c = 2
Which gives the coordinates of point D.
D(0, 1 , -1)
b) The area A of the parallelogram is given by (see formula in Scalar and Cross Products of 3D Vectors )
A = \( || \vec{AB} \times \vec{BC} || \)
Let us first calculate the cross product,
\( \vec{AB} \times \vec{BC} = {\begin{vmatrix}\vec{i}& \vec{j} &\vec{k} \\ -2 & -4 & 2 \\ -4 & -5 & -3 \end{vmatrix}} = {\begin{vmatrix} -4 & 2 \\ -5 & -3 \end{vmatrix}} \vec{i} - {\begin{vmatrix}-2 & 2\\ -4 & -3\end{vmatrix}} \vec{j} + {\begin{vmatrix}-2 & -4\\ -4& -5\end{vmatrix}} \vec{k} = 22\vec{i} + 14\vec{j} -6\vec{k} \)
then its magnitude which is the area A.
A = \( ||\vec{AB} \times \vec{BC}|| = \sqrt{22^2+14^2+(-6)^2} = 2\sqrt{179} \)
9) In the cube below find the angle between the diagonals AG and BH.
Solution
We first find the components of the vectors \( \vec{AG}\) and \( \vec{BH} \).
\( \vec{AG} = <2,2,2>\)
\( \vec{BH} = <-2,2,2> \)
Using the dot product, with θ being the angle between the vectors \( \vec{AG}\) and \( \vec{BH} \), we have (see question 5 above).
\( \cos \theta = \dfrac{\vec{AG} \cdot \vec{BG}}{||\vec{AG}|| || \vec{BH} ||} = \dfrac{2\cdot (-2)+2\cdot2+2\cdot2}{\sqrt{2^2+2^2+2^2}\sqrt{(-2)^2+2^2+2^2}} = \dfrac{1}{3}\)
\( \theta = \arccos \dfrac{1}{3} \approx 70.5^{\circ} \)
10) Find a vector that is orthogonal to the plane containing the points A(1,2,-3), B(0,-2,1) and C(-2,0,1).
Solution
A vector \( \vec{v} \) that is orthogonal to the plane defined above could be given by the cross product of any two distinct vectors in the plane. Hence.
\( \vec{v} = \vec{AB} \times \vec{AC} = {\begin{vmatrix}\vec{i}& \vec{j} &\vec{k} \\ -1 & -4 & 4 \\ -3 & -2 & 4 \end{vmatrix}} = {\begin{vmatrix} -4 & 4 \\ -2 & 4 \end{vmatrix}} \vec{i} - {\begin{vmatrix}-1 & 4\\ -3 & 4\end{vmatrix}} \vec{j} + {\begin{vmatrix}-1 & -4\\ -3& -2\end{vmatrix}} \vec{k} = -8\vec{i} - 8\vec{j} -10\vec{k} \)
Note that there are an infinite number of solutions.
11) Find the area of the triangle whose vertices are the points A(1,0,-3), B(1,-2,0) and C(0,2,1).
Solution
The area \( A_t \) of a triangle is given by half the magnitude of the cross product of any two vectors made by the sides of the triangle. Hence
\( A_t = \vec{AB} \times \vec{AC} = {\begin{vmatrix}\vec{i}& \vec{j} &\vec{k} \\ 0 & -2 & 3 \\ -1 & 2 & 4 \end{vmatrix}} = {\begin{vmatrix} -2 & 3 \\ 2 & 4 \end{vmatrix}} \vec{i} - {\begin{vmatrix}0 & 3\\ -1 & 4\end{vmatrix}} \vec{j} + {\begin{vmatrix}0 & -2\\ -1& 2\end{vmatrix}} \vec{k} = -14\vec{i} - 3\vec{j} -2\vec{k} \)
\( A_t = (1/2) || \vec{AB} \times \vec{AC} || = \sqrt{(-14)^2 + (-3)^2 + (-2)^2} = (1/2) \sqrt{209} \) unit 2
12)Find the volume of the parallelepiped shown below.
Solution
The volume V of the parallelepiped is given by
V \( = |\vec{u}\cdot (\vec{v} \times \vec{w})| = | \vec{v}\cdot (\vec{w} \times \vec{u})| = | \vec{w}\cdot (\vec{v} \times \vec{u})| \)
Let us first find the components of the vectors \( \vec{u} \), \( \vec{v} \) and \( \vec{w}\).
\( \vec{u} = <-3,0,7> \)
\( \vec{v} = <-8,0,0> \)
\( \vec{w} = <0,-9,0> \)
\( \vec{u}\cdot (\vec{v} \times \vec{w})= <-3 , 0 , 7> \cdot {\begin{vmatrix}\vec{i}& \vec{j} &\vec{k} \\ -8 & 0 & 0 \\ 0 & -9 & 0 \end{vmatrix}}\)
\( = <-3 , 0 , 7> \cdot \left\{ {\begin{vmatrix} 0 & 0 \\ -9 & 0 \end{vmatrix}} \vec{i} - {\begin{vmatrix}-8 & 0\\ 0 & 0\end{vmatrix}} \vec{j} + {\begin{vmatrix}-8 & 0\\ 0 & -9\end{vmatrix}} \vec{k} \right\} \)
\( = <-3 , 0 , 7> \cdot <0 , 0 , 72> = 0 + 0 + 7\cdot72 = 504\)
The volume is
\( V = |504| = 504 \) unit 3
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