9 The line l₁ and the plane π₁ have vector equations l₁:r = [−1, 2, 3] + λ[1, 2, −1] and π₁:r · [−2, 1, 3]…

Question

9 The line l₁ and the plane π₁ have vector equations l₁:r = [−1, 2, 3] + λ[1, 2, −1] and π₁:r · [−2, 1, 3] = 9, where λ is a real parameter. The point P has coordinates (4,2,−1). (a) Find the coordinates of the point of intersection, Q, of the line l₁ and the plane π₁.

ScanSolve AI Tutor · Accepted Answer

Here are the solutions for question 9a to 9d.

9a)
To find the point of intersection Q, substitute the parametric equations of line $l_1$ into the equation of plane $\pi_1$ .
The line $l_1$ is given by $\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ , so $x = 2 + \lambda$ , $y = -1 + 2\lambda$ , $z = 3 - \lambda$ .
The plane $\pi_1$ is given by $\mathbf{r} \cdot \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = 9$ .

Substitute the components of $\mathbf{r}$ from the line equation into the plane equation.
$(2 + \lambda)(2) + (-1 + 2\lambda)(1) + (3 - \lambda)(3) = 9$
Expand and solve for $\lambda$ .
$4 + 2\lambda - 1 + 2\lambda + 9 - 3\lambda = 9$ $12 + \lambda = 9$ $\lambda = 9 - 12$ $\lambda = -3$
Substitute $\lambda = -3$ back into the line equation to find the coordinates of Q.
$x = 2 + (-3) = -1$ $y = -1 + 2(-3) = -1 - 6 = -7$ $z = 3 - (-3) = 3 + 3 = 6$
The coordinates of Q are $\boxed{(-1, -7, 6)}$ .

9b)
To find the acute angle between the line $l_1$ and the plane $\pi_1$ , we use the formula $\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}$ , where $\mathbf{d}$ is the direction vector of the line and $\mathbf{n}$ is the normal vector of the plane.
The direction vector of $l_1$ is $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ .
The normal vector of $\pi_1$ is $\mathbf{n} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$ .
Calculate the dot product $\mathbf{d} \cdot \mathbf{n}$ .
$\mathbf{d} \cdot \mathbf{n} = (1)(2) + (2)(1) + (-1)(3) = 2 + 2 - 3 = 1$
Calculate the magnitudes of $\mathbf{d}$ and $\mathbf{n}$ .
$|\mathbf{d}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$ $|\mathbf{n}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$
Calculate $\sin \theta$ and find $\theta$ .
$\sin \theta = \frac{|1|}{\sqrt{6}\sqrt{14}} = \frac{1}{\sqrt{84}}$ $\theta = \arcsin\left(\frac{1}{\sqrt{84}}\right)$ $\theta \approx \arcsin(0.1091089) \approx 6.273^\circ$
Rounding to 1 decimal place, the acute angle is $\boxed{6.3^\circ}$ .

9c)
To find the perpendicular distance from point P $(4, 2, -1)$ to line $l_1$ , we can use the formula $D = \frac{|\vec{AP} \times \mathbf{d}|}{|\mathbf{d}|}$ , where A is a point on the line and $\mathbf{d}$ is the direction vector of the line.
A point on $l_1$ is $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ .
The direction vector of $l_1$ is $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ .
The point P is $\mathbf{p} = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}$ .
Find the vector $\vec{AP}$ .
$\vec{AP} = \mathbf{p} - \mathbf{a} = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix}$
Calculate the cross product $\vec{AP} \times \mathbf{d}$ .
$\vec{AP} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -4 \\ 1 & 2 & -1 \end{vmatrix} = \mathbf{i}((3)(-1) - (-4)(2)) - \mathbf{j}((2)(-1) - (-4)(1)) + \mathbf{k}((2)(2) - (3)(1))$ $= \mathbf{i}(-3 + 8) - \mathbf{j}(-2 + 4) + \mathbf{k}(4 - 3) = 5\mathbf{i} - 2\mathbf{j} + 1\mathbf{k} = \begin{pmatrix} 5 \\ -2 \\ 1 \end{pmatrix}$
Calculate the magnitudes $|\vec{AP} \times \mathbf{d}|$ and $|\mathbf{d}|$ .
$|\vec{AP} \times \mathbf{d}| = \sqrt{5^2 + (-2)^2 + 1^2} = \sqrt{25 + 4 + 1} = \sqrt{30}$ $|\mathbf{d}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$
Calculate the perpendicular distance.
$D = \frac{|\vec{AP} \times \mathbf{d}|}{|\mathbf{d}|} = \frac{\sqrt{30}}{\sqrt{6}} = \sqrt{\frac{30}{6}} = \sqrt{5}$
The perpendicular distance is $\boxed{\sqrt{5}}$ .

9d)
To find the Cartesian equation of plane $\pi_2$ which contains both line $l_1$ and point P, we need a normal vector to the plane and a point on the plane.
The line $l_1$ passes through point A $(2, -1, 3)$ and has direction vector $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ .
Point P is $(4, 2, -1)$ .
Find two non-parallel vectors lying in the plane.
One vector is the direction vector of the line: $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ .
Another vector can be formed by connecting a point on the line (A) to point P: $\vec{AP} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix}$ (calculated in part 9c).
Calculate the normal vector $\mathbf{n_2}$ to the plane by taking the cross product of these two vectors.
$\mathbf{n_2} = \mathbf{d} \times \vec{AP} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & 3 & -4 \end{vmatrix}$ $= \mathbf{i}((2)(-4) - (-1)(3)) - \mathbf{j}((1)(-4) - (-1)(2)) + \mathbf{k}((1)(3) - (2)(2))$ $= \mathbf{i}(-8 + 3) - \mathbf{j}(-4 + 2) + \mathbf{k}(3 - 4)$ $= -5\mathbf{i} + 2\mathbf{j} - 1\mathbf{k} = \begin{pmatrix} -5 \\ 2 \\ -1 \end{pmatrix}$
(Note: Using $\vec{AP} \times \mathbf{d}$ from part 9c gives $\begin{pmatrix} 5 \\ -2 \\ 1 \end{pmatrix}$ , which is just the negative of this normal vector. Both are valid normal vectors for the same plane.)
Write the Cartesian equation of the plane using the normal vector $\mathbf{n_2} = \begin{pmatrix} -5 \\ 2 \\ -1 \end{pmatrix}$ and a point on the plane (e.g., P $(4, 2, -1)$ ).
The equation is of the form $ax + by + cz = D$ .
$-5x + 2y - z = D$
Substitute P $(4, 2, -1)$ :
$-5(4) + 2(2) - (-1) = D$ $-20 + 4 + 1 = D$ $D = -15$
So the Cartesian equation is $-5x + 2y - z = -15$ .
Multiplying by -1 to make the leading coefficient positive (standard form):
$5x - 2y + z = 15$
The Cartesian equation of plane $\pi_2$ is $\boxed{5x - 2y + z = 15}$ .

9 The line l₁ and the plane π₁ have vector equations l₁:r = [-1, 2, 3] + λ[1, 2, -1] and π₁:r · [-2, 1, 3] = 9, where λ is a real parameter. The point P has coordinates (4,2,-1). (a) Find the coordinates of the point of intersection, Q, of the line l₁ and the plane π₁.

Step-by-step solution

Substitute the components of $\mathbf{r}$ from the line equation into the plane equation.

Expand and solve for $\lambda$ .

Substitute $\lambda = -3$ back into the line equation to find the coordinates of Q.

Calculate the dot product $\mathbf{d} \cdot \mathbf{n}$ .

Calculate the magnitudes of $\mathbf{d}$ and $\mathbf{n}$ .

Calculate $\sin \theta$ and find $\theta$ .

Find the vector $\vec{AP}$ .

Calculate the cross product $\vec{AP} \times \mathbf{d}$ .

Calculate the magnitudes $|\vec{AP} \times \mathbf{d}|$ and $|\mathbf{d}|$ .

Calculate the perpendicular distance.

Find two non-parallel vectors lying in the plane.

Calculate the normal vector $\mathbf{n_2}$ to the plane by taking the cross product of these two vectors.

Write the Cartesian equation of the plane using the normal vector $\mathbf{n_2} = \begin{pmatrix} -5 \\ 2 \\ -1 \end{pmatrix}$ and a point on the plane (e.g., P $(4, 2, -1)$ ).

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More Mathematics Questions

9 The line l₁ and the plane π₁ have vector equations l₁:r = [-1, 2, 3] + λ[1, 2, -1] and π₁:r · [-2, 1, 3] = 9, where λ is a real parameter. The point P has coordinates (4,2,-1). (a) Find the coordinates of the point of intersection, Q, of the line l₁ and the plane π₁.

Step-by-step solution

Substitute the components of r\mathbf{r}r from the line equation into the plane equation.

Expand and solve for λ\lambdaλ.

Substitute λ=−3\lambda = -3λ=−3 back into the line equation to find the coordinates of Q.

Calculate the dot product d⋅n\mathbf{d} \cdot \mathbf{n}d⋅n.

Calculate the magnitudes of d\mathbf{d}d and n\mathbf{n}n.

Calculate sin⁡θ\sin \thetasinθ and find θ\thetaθ.

Find the vector AP⃗\vec{AP}AP.

Calculate the cross product AP⃗×d\vec{AP} \times \mathbf{d}AP×d.

Calculate the magnitudes ∣AP⃗×d∣|\vec{AP} \times \mathbf{d}|∣AP×d∣ and ∣d∣|\mathbf{d}|∣d∣.