MAT1503
EXAM PACK
,LINEAR ALGEBRA 1 - MAT1503
ASSIGNMENT 4 AUGUST 2025
EXAM ANSWERS
Question 1
1.1 Find an equation for the plane passing through the origin and parallel to
\[ -x+3y-2z=6. \]
Parallel planes have the same normal vector.
The normal vector is
\[ \mathbf n=\langle-1,3,-2\rangle. \]
A plane through the origin has the form
\[ - x+3y-2z=d. \]
Since it passes through \((0,0,0)\),
\[ d=0. \]
Answer
\[ \boxed{-x+3y-2z=0} \]
1.2 Find the distance between \((-1,-2,0)\) and the plane
\[ 3x-y+4z=-2. \]
Write the plane as
,\[ 3x-y+4z+2=0. \]
The distance from \((x_0,y_0,z_0)\) to \(Ax+By+Cz+D=0\) is
\[ d=\frac{|Ax_0+By_0+Cz_0+D|} {\sqrt{A^2+B^2+C^2}}. \]
Thus
\[ d= \frac{|3(-1)-(-2)+4(0)+2|} {\sqrt{3^2+(-1)^2+4^2}}. \]\[ d=\frac{|-3+2+2|}{\sqrt{26}}
=\frac1{\sqrt{26}}. \]
Answer
\[ \boxed{d=\frac1{\sqrt{26}}=\frac{\sqrt{26}}{26}} \]
Question 2
The vectors and vector-calculus exercises are given in the assignment on pages 3–4.
2025 A4 Final - MAT1503 Linear … 2025 A4 Final - MAT1503 Linear …
2.1 Find the angle between
\[ \mathbf v_1=\langle-1,1,0,-1\rangle \]
and
\[ \mathbf v_2=\langle1,-1,3,-2\rangle. \]
Use
\[ \mathbf v_1\cdot\mathbf v_2 =|\mathbf v_1||\mathbf v_2|\cos\theta. \]
First,
\[ \mathbf v_1\cdot\mathbf v_2 =(-1)(1)+(1)(-1)+(0)(3)+(-1)(-2) \]\[ =-1-1+0+2=0. \]
Therefore,
, \[ \cos\theta=0 \]
and hence
\[ \boxed{\theta=90^\circ}. \]
Therefore, the vectors are perpendicular.
Answer
\[ \boxed{\theta=90^\circ,\qquad \mathbf v_1\perp\mathbf v_2} \]
They are neither parallel nor merely non-related; they are perpendicular.
2.2 Direction cosines and direction angles of
\[ \mathbf r=\left\langle0,-1,-2,\frac34\right\rangle. \]
Its magnitude is
\[ |\mathbf r| =\sqrt{0^2+(-1)^2+(-2)^2+\left(\frac34\right)^2}. \]\[ =\sqrt{1+4+\frac9{16}}
=\sqrt{\frac{89}{16}} =\frac{\sqrt{89}}4. \]
Therefore the direction cosines are
\[ \cos\alpha=\frac{0}{\sqrt{89}/4}=0, \]\[ \cos\beta=\frac{-1}{\sqrt{89}/4} =-
\frac4{\sqrt{89}}, \]\[ \cos\gamma=\frac{-2}{\sqrt{89}/4} =-\frac8{\sqrt{89}}, \]
and for the fourth coordinate,
\[ \cos\delta= \frac{3/4}{\sqrt{89}/4} =\frac3{\sqrt{89}}. \]
Direction cosines
\[ \boxed{ \left( 0,-\frac4{\sqrt{89}}, -\frac8{\sqrt{89}}, \frac3{\sqrt{89}} \right)} \]
The corresponding direction angles are
EXAM PACK
,LINEAR ALGEBRA 1 - MAT1503
ASSIGNMENT 4 AUGUST 2025
EXAM ANSWERS
Question 1
1.1 Find an equation for the plane passing through the origin and parallel to
\[ -x+3y-2z=6. \]
Parallel planes have the same normal vector.
The normal vector is
\[ \mathbf n=\langle-1,3,-2\rangle. \]
A plane through the origin has the form
\[ - x+3y-2z=d. \]
Since it passes through \((0,0,0)\),
\[ d=0. \]
Answer
\[ \boxed{-x+3y-2z=0} \]
1.2 Find the distance between \((-1,-2,0)\) and the plane
\[ 3x-y+4z=-2. \]
Write the plane as
,\[ 3x-y+4z+2=0. \]
The distance from \((x_0,y_0,z_0)\) to \(Ax+By+Cz+D=0\) is
\[ d=\frac{|Ax_0+By_0+Cz_0+D|} {\sqrt{A^2+B^2+C^2}}. \]
Thus
\[ d= \frac{|3(-1)-(-2)+4(0)+2|} {\sqrt{3^2+(-1)^2+4^2}}. \]\[ d=\frac{|-3+2+2|}{\sqrt{26}}
=\frac1{\sqrt{26}}. \]
Answer
\[ \boxed{d=\frac1{\sqrt{26}}=\frac{\sqrt{26}}{26}} \]
Question 2
The vectors and vector-calculus exercises are given in the assignment on pages 3–4.
2025 A4 Final - MAT1503 Linear … 2025 A4 Final - MAT1503 Linear …
2.1 Find the angle between
\[ \mathbf v_1=\langle-1,1,0,-1\rangle \]
and
\[ \mathbf v_2=\langle1,-1,3,-2\rangle. \]
Use
\[ \mathbf v_1\cdot\mathbf v_2 =|\mathbf v_1||\mathbf v_2|\cos\theta. \]
First,
\[ \mathbf v_1\cdot\mathbf v_2 =(-1)(1)+(1)(-1)+(0)(3)+(-1)(-2) \]\[ =-1-1+0+2=0. \]
Therefore,
, \[ \cos\theta=0 \]
and hence
\[ \boxed{\theta=90^\circ}. \]
Therefore, the vectors are perpendicular.
Answer
\[ \boxed{\theta=90^\circ,\qquad \mathbf v_1\perp\mathbf v_2} \]
They are neither parallel nor merely non-related; they are perpendicular.
2.2 Direction cosines and direction angles of
\[ \mathbf r=\left\langle0,-1,-2,\frac34\right\rangle. \]
Its magnitude is
\[ |\mathbf r| =\sqrt{0^2+(-1)^2+(-2)^2+\left(\frac34\right)^2}. \]\[ =\sqrt{1+4+\frac9{16}}
=\sqrt{\frac{89}{16}} =\frac{\sqrt{89}}4. \]
Therefore the direction cosines are
\[ \cos\alpha=\frac{0}{\sqrt{89}/4}=0, \]\[ \cos\beta=\frac{-1}{\sqrt{89}/4} =-
\frac4{\sqrt{89}}, \]\[ \cos\gamma=\frac{-2}{\sqrt{89}/4} =-\frac8{\sqrt{89}}, \]
and for the fourth coordinate,
\[ \cos\delta= \frac{3/4}{\sqrt{89}/4} =\frac3{\sqrt{89}}. \]
Direction cosines
\[ \boxed{ \left( 0,-\frac4{\sqrt{89}}, -\frac8{\sqrt{89}}, \frac3{\sqrt{89}} \right)} \]
The corresponding direction angles are
