Quiz about Quizizz 5.4 Applications of Vectors in Geometry

Question 1

Find the parametric equation of a straight line passing through the point (3, -1, 2) and is parallel to the vector i+2j-k

Accepted Answer

$$x=3+t,\ y=-1+2t,\ z=2-t$$

Answer

$$x-3=\frac{y+1}{2}=2-z$$

Answer

$$\frac{x-1}{2}=2-y=\frac{z-2}{3}$$

Answer

$$x=1+2t,\ y=2-t,\ z=2+3t$$

Question 2

Find the vector equation of a line that passes through points A(2, -3, 4) and B(1, 2, 7).

Accepted Answer

r = (2i-3j+4k) + t(-i+5j+3k)

Answer

r = (2i-3j+4k) + t(-2i+j-k)

Answer

r = (3i-2j+k) + t(-i+5j+3k)

Answer

r = (i+2j+7k) + t(2i-3j+4k)

Question 3

Find the vector equation of a line that passes through the point A(3, -4, 1) and is parallel to another line $$L_2:\ \frac{x-1}{2}=1-y=\frac{z-1}{3}$$

Accepted Answer

r = (3i-4j+k) + t(2i-j+3k)

Answer

r = (3i-2j+2k) + t(2i-j+3k)

Answer

r = (3i-4j+k) + t(3i-2j+2k)

Answer

r = (3i-2j+k) + t(2i-3j+4k)

Question 4

Find the angle between 2 straight lines if

$$L_1:$$ r=(3i+7j-k)+t(2i-3j+k)

$$L_2:$$ r=(2i-j+k)+s(3i-j-2k)

Accepted Answer

$$60\degree$$

Answer

$$47.5\degree$$

Answer

$$72.3\degree$$

Answer

$$30\degree$$

Question 5

Calculate the angle between $$L_1:\ \frac{x+1}{-2}=y-3=\frac{z-2}{-2}$$ and $$L_2:\ \frac{x-3}{6}=\frac{y+2}{2}=\frac{z+1}{-3}$$

Accepted Answer

$$101\degree$$

Answer

$$98\degree$$

Answer

$$113\degree$$

Answer

$$87\degree$$

Question 6

Given $$L_1:\ r=\left(2i+3j-k\right)+\alpha j$$ , $$L_2:\ r=-5k+\beta\left(i-\lambda j+2k\right)$$ and the angle between the 2 lines is $$143.3\degree$$ . Hence, find the value of $$\lambda$$ if $$\lambda>0$$ .

Hint: Use 3@4 d.p. in your calculations

Accepted Answer

3

Answer

2

Answer

4

Answer

5

Question 7

Find the equation of the plane that is orthogonal to the line $$x=4+t,\ y=1−2t,\ z=8t$$ and goes through the point $$P(3,2,1).$$

(Hint: orthogonal : The line is perpendicular to the plane)

Accepted Answer

$$x−2y+8z=7$$

Answer

$$x+2y+8z=7$$

Answer

$$3x−4y+8z=7$$

Answer

$$3x−4y+8z=-7$$

Question 8

Find the equation of the plane that is parallel to the plane $$5x−3y+2z=6$$ and goes through the point $$P(4,−1,2).$$

Accepted Answer

$$5x-3y+2z=27$$

Answer

$$4x+y-2z=11$$

Answer

$$4x−y+2z=23$$

Answer

$$5x+3y-2z=17$$

Question 9

Find the equation of the plane that contains the intersecting lines $$x=4+t_1,\ y=2t_1,\ z=1−3t_1$$ and $$x=4−3t_2,\ y=3t_2,\ z=1+2t_2.$$

Accepted Answer

$$13x+7y+9z=0$$

Answer

$$13x+7y+9z=-61$$

Answer

$$13x+7y-9z=61$$

Answer

$$13x-7y+9z=0$$

Question 10

What is the correct formula required to solve for the angle between 2 planes?

Accepted Answer

$$ heta=\cos^{-1}\ \frac{n_1.n_2}{\left|n_1 ight|\left|n_2 ight|}$$

Answer

$$ heta=\sin^{-1}\ \frac{n_1.n_2}{\left|n_1 ight|\left|n_2 ight|}$$

Answer

$$ heta=\cos^{-1}\ \frac{n_1.v_2}{\left|n_1 ight|\left|v_2 ight|}$$

Answer

$$ heta=\sin^{-1}\ \frac{n_1.v_2}{\left|n_1 ight|\left|v_2 ight|}$$

Question 11

Calculate the angle between the planes $$3x+y-4z=8$$

and $$4x-3y+z=-2$$

Accepted Answer

$$78.9\degree$$

Answer

$$102.5\degree$$

Answer

$$96.4\degree$$

Answer

$$87.2\degree$$

Question 12

In order to find the point of intersection between a line and a plane, you need

Accepted Answer

(1) Parametric eq. of the line

(2) Cartesian eq. of the plane

Answer

(1) Parametric eq. of the line

(2) Vector eq. of the plane

Answer

(1) Cartesian eq. of the line

(2) Parametric eq. of the plane

Answer

(1) Vector eq. of the line

(2) Parametric eq. of the plane

Question 13

Find point of intersection between line $$r=i-j+2k+t\left(2i+3j-k\right)$$ and plane $$r.\left(3i-j+2k\right)=11$$

Accepted Answer

$$\left(7,\ 8,\ -1\right)$$

Answer

$$\left(5,\ 2,\ -1\right)$$

Answer

$$\left(-3,\ 8,\ 2\right)$$

Answer

$$\left(7,\ -11,\ 3\right)$$

Question 14

If you have a line and a plane, define the angles $$\alpha$$ and $$ heta$$ .

Accepted Answer

(1) $$\alpha=$$ angle between vectors $$v$$ and $$n$$

(2) $$ heta=$$ angle between the line and the plane

Answer

(1) $$\alpha=$$ angle between the line and the plane

(2) $$ heta=$$ angle between vectors $$v$$ and $$n$$

Question 15

What is the correct formula required to solve for the angle between a line and a plane?

Accepted Answer

$$ heta=90\degree-\alpha$$

Answer

$$ heta=\cos^{-1}\ \frac{n.v}{\left|n ight|\left|v ight|}$$

Answer

$$ heta=90\degree+\alpha$$

Answer

$$ heta=\sin^{-1}\ \frac{n.v}{\left|n ight|\left|v ight|}$$

Question 16

Find the angle between the line $$r=i+2j-4k+m\left(2i-j+k\right)$$ and the plane $$13x-3j+11k=15$$

Accepted Answer

$$70.8\degree$$

Answer

$$23.2\degree$$

Answer

$$19.2\degree$$

Answer

$$66.8\degree$$

Quizizz 5.4 Applications of Vectors in Geometry

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