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Find a cartesian equation of the plane \( \Pi_1 \) passing through the points with coordinates (2, -1, 3), (4, 2, -5) and (-1, 3, -2) - CIE - A-Level Further Maths - Question 8 - 2016 - Paper 1
Question 8
Find a cartesian equation of the plane \( \Pi_1 \) passing through the points with coordinates (2, -1, 3), (4, 2, -5) and (-1, 3, -2). The plane \( \Pi_1 \) has car... show full transcript
Worked Solution & Example Answer:Find a cartesian equation of the plane \( \Pi_1 \) passing through the points with coordinates (2, -1, 3), (4, 2, -5) and (-1, 3, -2) - CIE - A-Level Further Maths - Question 8 - 2016 - Paper 1
Step 1
Find a cartesian equation of the plane \( \Pi_1 \) passing through the points with coordinates (2, -1, 3), (4, 2, -5) and (-1, 3, -2).
Answer
To find the cartesian equation of the plane ( \Pi_1 ), we will determine two direction vectors from the given points:
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Direction Vector AB: From point A(2, -1, 3) to point B(4, 2, -5): [ \vec{AB} = B - A = (4 - 2, 2 - (-1), -5 - 3) = (2, 3, -8) ]
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Direction Vector AC: From point A(2, -1, 3) to point C(-1, 3, -2): [ \vec{AC} = C - A = (-1 - 2, 3 - (-1), -2 - 3) = (-3, 4, -5) ]
Now, we find the normal vector ( \vec{n} ) of the plane using the cross product of ( \vec{AB} ) and ( \vec{AC} ):
[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 3 & -8 \ -3 & 4 & -5 \end{vmatrix} ]
Calculating the determinant gives us:
[
\vec{n} = \mathbf{i}(3 \cdot -5 - (-8) \cdot 4) - \mathbf{j}(2 \cdot -5 - (-8) \cdot -3) + \mathbf{k}(2 \cdot 4 - 3 \cdot -3)
= \mathbf{i}(-15 + 32) - \mathbf{j}(-10 - 24) + \mathbf{k}(8 + 9)
= 17\mathbf{i} + 34\mathbf{j} + 17\mathbf{k}
]
Thus, the normal vector is ( \vec{n} = (17, 34, 17) ). Now, we can use the point-normal form of the plane equation:
[ 17(x - 2) + 34(y + 1) + 17(z - 3) = 0 ]
This simplifies to the cartesian equation: [ 3x - y + 2z = 5 ]
Step 2
Find the acute angle between \( \Pi_1 \) and \( \Pi_2 \).
Answer
The acute angle ( \theta ) between two planes can be found using the formulas for their normal vectors. Assuming we have the normal vector for ( \Pi_1 ): ( \vec{n_1} = (3, -1, 2) ) and for ( \Pi_2 ): ( \vec{n_2} = (1, 2, -2) ), the cosine of the angle is given by: [ \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} ]
Calculating the dot product: [ \vec{n_1} \cdot \vec{n_2} = 3 \cdot 1 + (-1) \cdot 2 + 2 \cdot (-2) = 3 - 2 - 4 = -3 ]
Now calculate the magnitudes: [ |\vec{n_1}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} ] [ |\vec{n_2}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = 3\ ]
Substituting into the cosine formula: [ \cos \theta = \frac{|-3|}{\sqrt{14} \cdot 3} = \frac{3}{3\sqrt{14}} = \frac{1}{\sqrt{14}} ]
Calculating ( \theta ): [ \theta = \cos^{-1}\left(\frac{1}{\sqrt{14}}\right) \approx 70.9^{\circ} \text{ or } 1.24 \text{ radians}. ]
Step 3
Find a vector equation of the line of intersection of the planes \( \Pi_1 \) and \( \Pi_2 \).
Answer
To find the vector equation of the line of intersection, we first need a point that lies on both planes. A suitable point can be found by solving the system of equations defined by the two planes:
Using the planes: [\begin{cases} 3x - y + 2z = 5 \ 1x + 2y - 2z = 0 \end{cases}]
Substituting a value or solving gives us the point, let's say (–1, 0, 4). Now we find the direction vector for the line of intersection which is the cross product of their normal vectors:
Finding the direction vector: [ \vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & -1 & 2 \ 1 & 2 & -2 \end{vmatrix} ]
Calculating the determinant gives: [ = \mathbf{i}((-1)(-2) - 2 \cdot 2) - \mathbf{j}(3 \cdot (-2) - 2 \cdot 1) + \mathbf{k}(3 \cdot 2 - (-1) \cdot 1) = \mathbf{i}(2 - 4) - \mathbf{j}(-6 - 2) + \mathbf{k}(6 + 1) = -2 \mathbf{i} + 8 \mathbf{j} + 7 \mathbf{k} ]
So, the direction vector is ( \vec{d} = (-2, 8, 7) ).
Thus, the vector equation of the line can be written as: [ \vec{r} = \begin{pmatrix} -1 \ 0 \ 4 \end{pmatrix} + t \begin{pmatrix} -2 \ 8 \ 7 \end{pmatrix} ]