Free preview 2/15

Question 1
A line $l$ has parametric equations $x = 3 + 2 t$ , $y = - 1 + t$ , $z = 4 - t$ . A plane $Π$ has equation $3 x - y + 2 z = 15$ . Which of the following correctly describes the relationship between $l$ and $Π$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
CThe line intersects the plane at exactly one point
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Extract direction and normal vectors
The direction vector of the line is $v = (2, 1, - 1)$ and the normal to the plane is $n = (3, - 1, 2)$ .
Step 2: Compute the dot product $\mathbf{v} \cdot \mathbf{n}$
$v \cdot n = 2 (3) + 1 (- 1) + (- 1) (2) = 6 - 1 - 2 = 3 \neq = 0$ Since the dot product is non-zero, the line is not parallel to the plane.
Step 3: Conclude the intersection type
Because $v \cdot n \neq = 0$ , there is exactly one point of intersection. To verify, substitute: $3 (3 + 2 t) - (- 1 + t) + 2 (4 - t) = 15 \Rightarrow 9 + 6 t + 1 - t + 8 - 2 t = 15 \Rightarrow 18 + 3 t = 15 \Rightarrow t = - 1$ , giving a unique solution.
Method #2Approach 2
Step 1: Identify what determines the relationship
The key quantity is $v \cdot n$ . If it is zero, the line is parallel to or in the plane; if non-zero, there is a unique intersection.
Step 2: Eliminate 'parallel and does not intersect'
This requires $v \cdot n = 0$ . But $2 (3) + 1 (- 1) + (- 1) (2) = 3 \neq = 0$ , so this option is incorrect.
Step 3: Eliminate 'lies entirely within the plane'
This also requires $v \cdot n = 0$ , which fails. Additionally, checking the point $(3, - 1, 4)$ : $3 (3) - (- 1) + 2 (4) = 9 + 1 + 8 = 18 \neq = 15$ , so the line is not in the plane.
Step 4: Eliminate 'perpendicular to the plane'
A line perpendicular to a plane would have its direction vector parallel to the normal, i.e. $v = k n$ . Clearly $(2, 1, - 1)$ is not a scalar multiple of $(3, - 1, 2)$ , so this is incorrect.
Step 5: Select the correct answer
Since $v \cdot n = 3 \neq = 0$ , the line crosses the plane at exactly one point. The correct answer is 'The line intersects the plane at exactly one point'.
Question 2
A line $l$ passes through the points $A (1, - 2, 3)$ and $B (4, 0, 1)$ . A plane $Π$ has equation $3 x + 6 y - 3 z = 9$ . Which statement about $l$ and $Π$ is correct?
No clue? Show me the answer
Correct answerCorrect!Incorrect
BThe line lies entirely within the plane
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Find direction and normal vectors
Direction vector: $v = B - A = (3, 2, - 2)$ . Normal vector: $n = (3, 6, - 3)$ , or equivalently $(1, 2, - 1)$ .
Step 2: Compute $\mathbf{v} \cdot \mathbf{n}$
$v \cdot n = 3 (3) + 2 (6) + (- 2) (- 3) = 9 + 12 + 6 = 27$ Wait — let me recheck. Using simplified normal $(1, 2, - 1)$ : $3 (1) + 2 (2) + (- 2) (- 1) = 3 + 4 + 2 = 9 \neq = 0$ .
Step 3: Re-examine the plane equation
Dividing $3 x + 6 y - 3 z = 9$ by 3 gives $x + 2 y - z = 3$ , so $n = (1, 2, - 1)$ , $D = 3$ . Check $A (1, - 2, 3)$ : $1 + 2 (- 2) - 3 = 1 - 4 - 3 = - 6 \neq = 3$ . Hmm — let me reconsider by checking both points directly.
Step 4: Check both points in the original plane equation
Point $A (1, - 2, 3)$ : $3 (1) + 6 (- 2) - 3 (3) = 3 - 12 - 9 = - 18 \neq = 9$ . Point $B (4, 0, 1)$ : $3 (4) + 6 (0) - 3 (1) = 12 - 3 = 9$ ✓. Since $A$ is not on the plane but $B$ is, the line crosses the plane at exactly $B$ .
Step 5: Correct the answer
Since $v \cdot n = 9 \neq = 0$ , and substituting the parametric form yields a unique $t$ , the line meets the plane at exactly one point, which is $B (4, 0, 1)$ . The correct answer is 'The line meets the plane at the point $(4, 0, 1)$ only'.
Method #2Approach 2
Step 1: Set up parametric form and substitute
Write $l$ parametrically: $x = 1 + 3 t$ , $y = - 2 + 2 t$ , $z = 3 - 2 t$ . Substitute into $3 x + 6 y - 3 z = 9$ .
Step 2: Substitute and solve for $t$
$3 (1 + 3 t) + 6 (- 2 + 2 t) - 3 (3 - 2 t) = 9 \Rightarrow 3 + 9 t - 12 + 12 t - 9 + 6 t = 9 \Rightarrow 27 t - 18 = 9 \Rightarrow t = 1$ . This gives $x = 4, y = 0, z = 1$ , which is point $B$ .
Step 3: Eliminate 'lies entirely within the plane' and 'parallel'
Since we found a unique $t = 1$ , the line does not lie in the plane (not infinitely many solutions) and is not parallel (not zero solutions). Both these options are eliminated.
Step 4: Eliminate 'meets at a unique point other than $A$ or $B$'
The unique intersection point is $(4, 0, 1) = B$ , which IS one of the given points. So this option is incorrect.
Step 5: Select the correct answer
The line meets the plane at exactly the point $B (4, 0, 1)$ . The correct answer is 'The line meets the plane at the point $(4, 0, 1)$ only'.

Free preview

13 more questions in this topic

Free preview 2/15

Question 1
A line $l$ has parametric equations $x = 3 + 2 t$ , $y = - 1 + t$ , $z = 4 - t$ . A plane $Π$ has equation $3 x - y + 2 z = 15$ . Which of the following correctly describes the relationship between $l$ and $Π$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
CThe line intersects the plane at exactly one point
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Extract direction and normal vectors
The direction vector of the line is $v = (2, 1, - 1)$ and the normal to the plane is $n = (3, - 1, 2)$ .
Step 2: Compute the dot product $\mathbf{v} \cdot \mathbf{n}$
$v \cdot n = 2 (3) + 1 (- 1) + (- 1) (2) = 6 - 1 - 2 = 3 \neq = 0$ Since the dot product is non-zero, the line is not parallel to the plane.
Step 3: Conclude the intersection type
Because $v \cdot n \neq = 0$ , there is exactly one point of intersection. To verify, substitute: $3 (3 + 2 t) - (- 1 + t) + 2 (4 - t) = 15 \Rightarrow 9 + 6 t + 1 - t + 8 - 2 t = 15 \Rightarrow 18 + 3 t = 15 \Rightarrow t = - 1$ , giving a unique solution.
Method #2Approach 2
Step 1: Identify what determines the relationship
The key quantity is $v \cdot n$ . If it is zero, the line is parallel to or in the plane; if non-zero, there is a unique intersection.
Step 2: Eliminate 'parallel and does not intersect'
This requires $v \cdot n = 0$ . But $2 (3) + 1 (- 1) + (- 1) (2) = 3 \neq = 0$ , so this option is incorrect.
Step 3: Eliminate 'lies entirely within the plane'
This also requires $v \cdot n = 0$ , which fails. Additionally, checking the point $(3, - 1, 4)$ : $3 (3) - (- 1) + 2 (4) = 9 + 1 + 8 = 18 \neq = 15$ , so the line is not in the plane.
Step 4: Eliminate 'perpendicular to the plane'
A line perpendicular to a plane would have its direction vector parallel to the normal, i.e. $v = k n$ . Clearly $(2, 1, - 1)$ is not a scalar multiple of $(3, - 1, 2)$ , so this is incorrect.
Step 5: Select the correct answer
Since $v \cdot n = 3 \neq = 0$ , the line crosses the plane at exactly one point. The correct answer is 'The line intersects the plane at exactly one point'.
Question 2
A line $l$ passes through the points $A (1, - 2, 3)$ and $B (4, 0, 1)$ . A plane $Π$ has equation $3 x + 6 y - 3 z = 9$ . Which statement about $l$ and $Π$ is correct?
No clue? Show me the answer
Correct answerCorrect!Incorrect
BThe line lies entirely within the plane
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Find direction and normal vectors
Direction vector: $v = B - A = (3, 2, - 2)$ . Normal vector: $n = (3, 6, - 3)$ , or equivalently $(1, 2, - 1)$ .
Step 2: Compute $\mathbf{v} \cdot \mathbf{n}$
$v \cdot n = 3 (3) + 2 (6) + (- 2) (- 3) = 9 + 12 + 6 = 27$ Wait — let me recheck. Using simplified normal $(1, 2, - 1)$ : $3 (1) + 2 (2) + (- 2) (- 1) = 3 + 4 + 2 = 9 \neq = 0$ .
Step 3: Re-examine the plane equation
Dividing $3 x + 6 y - 3 z = 9$ by 3 gives $x + 2 y - z = 3$ , so $n = (1, 2, - 1)$ , $D = 3$ . Check $A (1, - 2, 3)$ : $1 + 2 (- 2) - 3 = 1 - 4 - 3 = - 6 \neq = 3$ . Hmm — let me reconsider by checking both points directly.
Step 4: Check both points in the original plane equation
Point $A (1, - 2, 3)$ : $3 (1) + 6 (- 2) - 3 (3) = 3 - 12 - 9 = - 18 \neq = 9$ . Point $B (4, 0, 1)$ : $3 (4) + 6 (0) - 3 (1) = 12 - 3 = 9$ ✓. Since $A$ is not on the plane but $B$ is, the line crosses the plane at exactly $B$ .
Step 5: Correct the answer
Since $v \cdot n = 9 \neq = 0$ , and substituting the parametric form yields a unique $t$ , the line meets the plane at exactly one point, which is $B (4, 0, 1)$ . The correct answer is 'The line meets the plane at the point $(4, 0, 1)$ only'.
Method #2Approach 2
Step 1: Set up parametric form and substitute
Write $l$ parametrically: $x = 1 + 3 t$ , $y = - 2 + 2 t$ , $z = 3 - 2 t$ . Substitute into $3 x + 6 y - 3 z = 9$ .
Step 2: Substitute and solve for $t$
$3 (1 + 3 t) + 6 (- 2 + 2 t) - 3 (3 - 2 t) = 9 \Rightarrow 3 + 9 t - 12 + 12 t - 9 + 6 t = 9 \Rightarrow 27 t - 18 = 9 \Rightarrow t = 1$ . This gives $x = 4, y = 0, z = 1$ , which is point $B$ .
Step 3: Eliminate 'lies entirely within the plane' and 'parallel'
Since we found a unique $t = 1$ , the line does not lie in the plane (not infinitely many solutions) and is not parallel (not zero solutions). Both these options are eliminated.
Step 4: Eliminate 'meets at a unique point other than $A$ or $B$'
The unique intersection point is $(4, 0, 1) = B$ , which IS one of the given points. So this option is incorrect.
Step 5: Select the correct answer
The line meets the plane at exactly the point $B (4, 0, 1)$ . The correct answer is 'The line meets the plane at the point $(4, 0, 1)$ only'.

Free preview

Step-by-step walkthrough

Step 1: Extract direction and normal vectors

Step 2: Compute the dot product $\mathbf{v} \cdot \mathbf{n}$

Step 3: Conclude the intersection type

Step 1: Identify what determines the relationship

Step 2: Eliminate 'parallel and does not intersect'

Step 3: Eliminate 'lies entirely within the plane'

Step 4: Eliminate 'perpendicular to the plane'

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Find direction and normal vectors

Step 2: Compute $\mathbf{v} \cdot \mathbf{n}$

Step 3: Re-examine the plane equation

Step 4: Check both points in the original plane equation

Step 5: Correct the answer

Step 1: Set up parametric form and substitute

Step 2: Substitute and solve for $t$

Step 3: Eliminate 'lies entirely within the plane' and 'parallel'

Step 4: Eliminate 'meets at a unique point other than $A$ or $B$'

Step 5: Select the correct answer

13 more questions in this topic

Step-by-step walkthrough

Step 1: Extract direction and normal vectors

Step 2: Compute the dot product $\mathbf{v} \cdot \mathbf{n}$

Step 3: Conclude the intersection type

Step 1: Identify what determines the relationship

Step 2: Eliminate 'parallel and does not intersect'

Step 3: Eliminate 'lies entirely within the plane'

Step 4: Eliminate 'perpendicular to the plane'

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Find direction and normal vectors

Step 2: Compute $\mathbf{v} \cdot \mathbf{n}$

Step 3: Re-examine the plane equation

Step 4: Check both points in the original plane equation

Step 5: Correct the answer

Step 1: Set up parametric form and substitute

Step 2: Substitute and solve for $t$

Step 3: Eliminate 'lies entirely within the plane' and 'parallel'

Step 4: Eliminate 'meets at a unique point other than $A$ or $B$'

Step 5: Select the correct answer

13 more questions in this topic