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The plane with equation \(x+y+z=a\), where \(a \in R\), intersects the coordinate axes at three points that form the vertices of a triangle.
The area of this triangle is given by
\(B\)
\(\text{Plane:}\ \ x+y+z=a\)
\(\text{Axis intercepts:}\ \ (a, 0,0),(0, a, 0),(0,0, a)\)
\(\text{Distance between \((a, 0,0)\) and \((0, a, 0)\) :}\)
\(d=\sqrt{(a-0)^2+(0-a)^2+(0-0)^2}=\sqrt{2} a\)
\(\text{Similarly for the other two sides}\)
\(\Rightarrow \ \text{Equilateral triangle with side length} \ \sqrt{2} a\)
\(A=\dfrac{1}{2} a b \sin C=\dfrac{1}{2}(\sqrt{2} a)^2 \times \sin 60^{\circ}=\dfrac{1}{2} \times 2 a^2 \times \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3} a^2}{2}\)
\(\Rightarrow B\)
Consider the two planes described by the equations \(2 x+2 y+z=2\) and \(a x+4 z=1\), where \(a\) is a positive constant.
The angle between the two planes is \(\cos ^{-1}\left(\dfrac{2}{3}\right)\).
The value of \(a\) satisfies the equation
\(A\)
\(\text{Plane 1:}\ \ 2 x+2 y+z=2\)
\(\text{Plane 2:}\ \ a x+4 z=1\)
\(\text{Angle between planes = angle between normals}\)
\(\displaystyle {\underset{\sim}{n}}_1=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right), \quad {\underset{\sim}{n}}_2=\left(\begin{array}{l}a \\ 0 \\ 4\end{array}\right)\)
| \({\underset{\sim}{n}}_1 \cdot {\underset{\sim}{n}}_2\) | \(=\abs{{\underset{\sim}{n}}_1}\abs{{\underset{\sim}{n}}_2}\cos(\theta)\) |
| \(2 a+4\) | \(=\sqrt{4+4+1} \times \sqrt{a^2+16} \times \dfrac{2}{3}\) |
| \(2 a+4\) | \(=3 \sqrt{a^2+16} \times \dfrac{2}{3}\) |
| \(a+2\) | \(=\sqrt{a^2+16}\) |
\(\Rightarrow A\)
Consider the points \(A(1,-2,3)\) and \(B(2,-5,-1)\). --- 3 WORK AREA LINES (style=lined) --- --- 14 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. \(\underset{\sim}{r}(t)=\underset{\sim}{i}-2 \underset{\sim}{j}+3 \underset{\sim}{k}+t(\underset{\sim}{i}-3\underset{\sim}{j}-4 \underset{\sim}{k}), \quad(t \in R)\) \(\text{or}\) \(\underset{\sim}{r}(t)=2\underset{\sim}{i}-5 \underset{\sim}{j}-\underset{\sim}{k}+t(\underset{\sim}{i}-3\underset{\sim}{j}-4 \underset{\sim}{k}), \quad (t \in R)\) b. \(\text{Shortest distance}=\dfrac{5 \sqrt{66}}{6}\) c. \(40x+12 y+z=19\) d.i. \(X(6,0,0), Y(0,-4,0), Z(0,0,3)\) d.ii. \(\text{Area}=3 \sqrt{29}\) a. \(\overrightarrow{AB}=\left(\begin{array}{c}2 \\ -5 \\ -1\end{array}\right)-\left(\begin{array}{c}1 \\ -2 \\ 3\end{array}\right)=\left(\begin{array}{c}1 \\ -3 \\ -4\end{array}\right)\) \(\underset{\sim}{r}(t)=\underset{\sim}{i}-2 \underset{\sim}{j}+3 \underset{\sim}{k}+t(\underset{\sim}{i}-3\underset{\sim}{j}-4 \underset{\sim}{k}), \quad(t \in R)\) \(\text{or}\) \(\underset{\sim}{r}(t)=2\underset{\sim}{i}-5 \underset{\sim}{j}-\underset{\sim}{k}+t(\underset{\sim}{i}-3\underset{\sim}{j}-4 \underset{\sim}{k}), \quad (t \in R)\) b. \({\underset{\sim}{r}}_\text{A}=\underset{\sim}{i}-2\underset{\sim}{j}+3 \underset{\sim}{k}\) \({\underset{\sim}{r}}_1=(2-t) \underset{\sim}{i}+(1+2 t) \underset{\sim}{j}+(t-3) \underset{\sim}{k}\) \({\underset{\sim}{r}}_1-{\underset{\sim}{r}}_\text{A}=(1-t)\underset{\sim}{i}+(2 t+3) \underset{\sim}{j}+(t-6) \underset{\sim}{k}\) \(\abs{{\underset{\sim}{r}}_1-{\underset{\sim}{r}}_\text{A}}=\sqrt{(1-t)^2+(2 t+3)^2+(t-6)^2}\) \(\text{Find \(t\) when} \ \ \dfrac{d}{dt}\left(\abs{{\underset{\sim}{r}}_1-{\underset{\sim}{r}}_\text{A}}\right)=0:\) \(t=\dfrac{1}{6}\) \(\therefore\ \text{Shortest distance}=\dfrac{5 \sqrt{66}}{6}\) c. \(C(0,2,-5)\) \(\text{Plane equation:} \ \dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\) \(\text{Solve:} \ \ \dfrac{1}{a}-\dfrac{2}{b}+\dfrac{3}{c}=1, \quad \dfrac{2}{a}-\dfrac{5}{b}-\dfrac{1}{c}=1, \quad \dfrac{0}{a}+\dfrac{2}{b}-\dfrac{5}{c}=1\) \(a=\dfrac{19}{40}, \quad b=\dfrac{19}{12}, \quad c=19\) \(\text{Equation of plane:}\) \(\dfrac{40x}{19}+\dfrac{12y}{19}+\dfrac{z}{19}=1 \ \Rightarrow \ 40x+12 y+z=19\) d.i. \(2 x-3 y+4 z=12\) d.ii \(\overrightarrow{XY}=\left(\begin{array}{c}0 \\ -4 \\ 0\end{array}\right)-\left(\begin{array}{l}6 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{c}-6 \\ -4 \\ 0\end{array}\right)\)
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\(2x=12\ \)
\(\ \Rightarrow \ x=6\ \)
\(\Rightarrow\ X(6,0,0)\)
\(-3 y=12\ \)
\(\ \Rightarrow \ y=-4\ \)
\(\Rightarrow\ Y(0,-4,0)\)
\(4 z=12\ \)
\(\ \Rightarrow \ z=3\ \)
\(\Rightarrow\ Z(0,0,3)\)
\(\overrightarrow{X Z}=\left(\begin{array}{l}0 \\ 0 \\ 3\end{array}\right)-\left(\begin{array}{l}6 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{c}-6 \\ 0 \\ 3\end{array}\right)\)
\(\text{Area}=\dfrac{1}{2}\abs{\overrightarrow{XY} \times \overrightarrow{XZ}}=\dfrac{1}{2}\abs{-12\underset{\sim}{i}+18 \underset{\sim}{j}-24 \underset{\sim}{k}}=\sqrt{261}=3 \sqrt{29}\)
The point of intersection of the line \(\underset{\sim}{ r }=\underset{\sim}{ i }+\underset{\sim}{ j }-2 \underset{\sim}{ k }+t(-2 \underset{\sim}{ i }+\underset{\sim}{ j }+3 \underset{\sim}{ k })\), where \(t \in R\) and the plane \(3 x-2 y+4 z=5\) is
\(D\)
\(\text{Find \(t\) such that \(\underset{\sim}{r}\) is a point on the plane.}\)
\(\text{Plane equation} \ \ 3x-2 y+4z=5\)
\(\text{Substitute} \ \ x=1-2t, \ y=1+t, \ z=-2+3t\)
\(\text{Solve for} \ t:\)
\(3(1-2 t)-2(1+t)+4(-2+3 t)=5\)
\(t=3\)
\(\text{Substitute \(t=3\) into \(\underset{\sim}{r}\):}\)
| \(\underset{\sim}{r}(3)\) | \(=(1-6,1+3,-2+9)\) |
| \(=(-5, 4, 7)\) |
\(\Rightarrow D\)