Problem 46 - Essential Calculus Early Transcendentals

Question

(a) Let \(P\) be a point not on the plane that passes through the points \(Q, R,\) and \(S .\) Show that the distance \(d\) from \(P\) to the plane is $$\begin{aligned} d &=\frac{|\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})|}{|\mathbf{a} \times \mathbf{b}|} \\ \text { where a }=\vec{Q R}, \mathbf{b} &=\vec{Q S}, \text { and } \mathbf{c}=\vec{Q P} .\end{aligned}$$ (b) Use the formula in part (a) to find the distance from the point \(P(2,1,4)\) to the plane through the points \(Q(1,0,0),\) \(R(0,2,0),\) and \(S(0,0,3).\)

Step-by-Step Solution

Answer

The distance from point \( P \) to the plane is \( \frac{17}{7} \).

Step 1: Determine Vector Representations

Calculate the vectors \( \mathbf{a} = \overrightarrow{QR} \), \( \mathbf{b} = \overrightarrow{QS} \), and \( \mathbf{c} = \overrightarrow{QP} \). For \( Q(1,0,0), R(0,2,0), S(0,0,3), P(2,1,4) \), we have:- \( \mathbf{a} = \langle 0-1, 2-0, 0-0 \rangle = \langle -1, 2, 0 \rangle \)- \( \mathbf{b} = \langle 0-1, 0-0, 3-0 \rangle = \langle -1, 0, 3 \rangle \)- \( \mathbf{c} = \langle 2-1, 1-0, 4-0 \rangle = \langle 1, 1, 4 \rangle \)

Step 2: Calculate the Cross Product \( \mathbf{b} \times \mathbf{c} \)

To find \( \mathbf{b} \times \mathbf{c} \), use the determinant formula:\[\mathbf{b} \times \mathbf{c} =\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-1 & 0 & 3 \1 & 1 & 4 \end{vmatrix}= \mathbf{i}(0 - 3) - \mathbf{j}(-4 - 3) + \mathbf{k}(-1 - 0)= -3\mathbf{i} + 7\mathbf{j} - 1\mathbf{k} = \langle -3, 7, -1 \rangle \]

Step 3: Calculate Dot Product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \)

Compute the dot product of \( \mathbf{a} \) and the calculated cross product:\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \langle -1, 2, 0 \rangle \cdot \langle -3, 7, -1 \rangle = (-1)(-3) + (2)(7) + (0)(-1) = 3 + 14 + 0 = 17\]

Step 4: Calculate Magnitude of \( \mathbf{a} \times \mathbf{b} \)

Now find the magnitude of the vector \( \mathbf{a} \times \mathbf{b} \), calculated similarly:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-1 & 2 & 0 \-1 & 0 & 3\end{vmatrix} = \mathbf{i}(2*3 - 0) - \mathbf{j}(-3 - 0) + \mathbf{k}(-1*0 - (-1)*2)= 6\mathbf{i} + 3\mathbf{j} + 2\mathbf{k} = \langle 6, 3, 2 \rangle \]The magnitude is \[ |\mathbf{a} \times \mathbf{b}| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{49} = 7\]

Step 5: Calculate Distance using Formula

Using the formula\[d = \frac{|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|}{|\mathbf{a} \times \mathbf{b}|}\]Plug in the values:\[d = \frac{|17|}{7} = \frac{17}{7} \]Therefore, the distance \( d \) from the point to the plane is \( \frac{17}{7} \).

Key Concepts

Understanding Vector Cross Product

The vector cross product is a fundamental operation in vector algebra. It results in a vector that is perpendicular (orthogonal) to both of the original vectors. The resulting cross product vector points in the direction given by the right-hand rule. This means, if you point your index finger in the direction of the first vector, and your middle finger in the direction of the second vector, your thumb points in the direction of the cross product.

The mathematical formula for computing a cross product of two vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) is given by the determinant of a 3x3 matrix:

• \[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} = \mathbf{i}(u_2v_3 - u_3v_2) - \mathbf{j}(u_1v_3 - u_3v_1) + \mathbf{k}(u_1v_2 - u_2v_1) \]

Cross products are extremely useful for finding a vector normal to a plane, as we can see in the problem above, where \( \mathbf{b} \times \mathbf{c} \) gives us the vector perpendicular to the plane defined by vectors \( \mathbf{b} \) and \( \mathbf{c} \).

Dot Product and Its Role

The dot product, also called the scalar product, of two vectors produces a scalar quantity. It reflects how much one vector extends in the direction of another vector. The dot product is calculated by multiplying the corresponding components of two vectors and then summing those products.

For vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), the dot product is expressed as:

• \[ \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \]

The dot product is crucial when determining how much one vector aligns with another, as in the exercise above, where \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) helps to derive the parallel component of the vector in determining the distance from a point to a plane. The absolute value of the dot product here represents a component of the area when intersected with another plane, contributing to the distance calculation formula.

Calculating the Magnitude of a Vector

The magnitude of a vector is essentially its length or size. It is calculated using the Pythagorean theorem extended into three dimensions. Finding the magnitude of a vector is important because it allows us to understand the scale of the vector in relation to other vectors.

For a vector \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \), the magnitude \( |\mathbf{u}| \) is given by:

• \[ |\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2} \]

In the context of the exercise provided, calculating the magnitude of the cross product \( \mathbf{a} \times \mathbf{b} \) is necessary to determine the denominator in the distance formula. This magnitude plays a crucial role because it normalizes the result, ensuring the distance is a scalar value that makes intuitive sense geometrically.