What value of \(k\), where \(k \in R\), will make the following planes perpendicular?
\(\Pi_1: \ 2 x-k y+3 z=1\)
\(\Pi_2: \ 2 k x+3 y-2 z=4\)
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What value of \(k\), where \(k \in R\), will make the following planes perpendicular?
\(\Pi_1: \ 2 x-k y+3 z=1\)
\(\Pi_2: \ 2 k x+3 y-2 z=4\)
\(C\)
\(\underset{\sim}{p_1} = 2\underset{\sim}{i}-k\underset{\sim}{j}+3\underset{\sim}{k}\)
\(\underset{\sim}{p_2} = 2k\underset{\sim}{i}+3\underset{\sim}{j}-2\underset{\sim}{k}\)
\(\text{Solve}\ \ \underset{\sim}{p_1} \cdot \underset{\sim}{p_2}=0\ \ \text{for}\ k\ \text{(by calc)}: \)
\(4k-3k-6=0\ \ \Rightarrow \ k=6\)
\(\Rightarrow C\)
A plane contains the points \( A(1,3,-2), B(-1,-2,4)\) and \( C(a,-1,5)\), where \(a\) is a real constant. The plane has a \(y\)-axis intercept of 2 at the point \(D\).
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a. \(D(0,2,0)\)
b. \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)
\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)
c. \(4 x+2 y+3 z =4\)
d. \(a=-\dfrac{9}{4}\)
e. \(A=\sqrt{29}\)
a. \(D(0,2,0)\)
b. \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)
\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)
c. \(\overrightarrow{A B} \times \overrightarrow{A D}=\left|\begin{array}{ccc}
\underset{\sim}{\text{i}} & \underset{\sim}{\text{j}} & \underset{\sim}{\text{k}} \\ -2 & -5 & 6 \\ -1 & -1 & 2\end{array}\right|\)
\(\ \quad \quad \quad \quad \ \begin{aligned} & =(-10+6)\underset{\sim}{\text{i}}-(-4+6)\underset{\sim}{\text{j}}+(2-5) \underset{\sim}{\text{k}} \\ & =-4 \underset{\sim}{i}-2 \underset{\sim}{j}-3 \underset{\sim}{k}\end{aligned}\)
\(\text{Plane:}\ \ -4 x-2 y-3 z=k\)
\(\text{Substitute}\ D(0,2,0)\ \text{into equation}\ \ \Rightarrow \ \text{k}=-4\)
\begin{aligned}
\therefore-4 x-2 y-3 z & =-4 \\
4 x+2 y+3 z & =4
\end{aligned}
d. \(\text{Find}\ a\ \Rightarrow \ \text{substitute}\ (a, 1,-5)\ \text{into plane equation:}\)
\begin{aligned}
-4 & =-4a+2-15 \\
4 a & =-9 \\
a & =-\dfrac{9}{4}
\end{aligned}
e. | \(\text{Area}\) | \(=|\overrightarrow{A B} \times \overrightarrow{A D}|\) |
\(=|-4\underset{\sim}{i}-2\underset{\sim}{j}-3\underset{\sim}{k}|\) | ||
\(=\sqrt{16+4+9} \) | ||
\(=\sqrt{29}\) |