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The plane \( \Pi_1 \) has equation \( \mathbf{r} = \frac{2}{3} \mathbf{i} + s \mathbf{j} + t \left( \mathbf{k} + \frac{-1}{2} \mathbf{j} \right) \) - CIE - A-Level Further Maths - Question 8 - 2013 - Paper 1
Question 8
The plane \( \Pi_1 \) has equation \( \mathbf{r} = \frac{2}{3} \mathbf{i} + s \mathbf{j} + t \left( \mathbf{k} + \frac{-1}{2} \mathbf{j} \right) \). Find a Cartesian... show full transcript
Worked Solution & Example Answer:The plane \( \Pi_1 \) has equation \( \mathbf{r} = \frac{2}{3} \mathbf{i} + s \mathbf{j} + t \left( \mathbf{k} + \frac{-1}{2} \mathbf{j} \right) \) - CIE - A-Level Further Maths - Question 8 - 2013 - Paper 1
Step 1
Find a Cartesian equation of \( \Pi_1 \)
Answer
To find the Cartesian equation of the plane ( \Pi_1 ) represented by the given vector equation, we first identify the normal vector by taking the coefficients of ( \mathbf{j} ) and ( \mathbf{k} ) from the given equation. The normal vector is ( \mathbf{n_1} = \mathbf{i} + 3\mathbf{j} - \mathbf{k} ) (from ( (1, 3, -1) )). We can represent the equation of the plane in the form: ( ax + by + cz = d ), where ( a, b, c ) are the components of the normal vector.
Using the point ( \left( \frac{2}{3}, 0, -1 \right) ) which lies on the plane, we substitute these values to find ( d ).
Thus, the Cartesian equation of ( \Pi_1 ) is: [ x + 3y - z = 12. ]
Step 2
Find the acute angle between \( \Pi_1 \) and \( \Pi_2 \)
Answer
To find the acute angle between the two planes, we need the normals:\n1. Normal of ( \Pi_1 ) is ( \mathbf{n_1} = (1, 3, -1) )\n2. Normal of ( \Pi_2 ) is ( \mathbf{n_2} = (2, -1, 1) )\n\nThe cosine of the angle ( \theta ) between the two planes can be found using the formula:
[ \cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}| |\mathbf{n_2}|} ]\n
Calculating the dot product: ( \mathbf{n_1} \cdot \mathbf{n_2} = 1 \cdot 2 + 3 \cdot (-1) + (-1) \cdot 1 = 2 - 3 - 1 = -2 )\n
Calculating the magnitudes:\n1. ( | \mathbf{n_1} | = \sqrt{1^2 + 3^2 + (-1)^2} = \sqrt{11} )\n2. ( | \mathbf{n_2} | = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6} )\n\nSo, ( \cos \theta = \frac{-2}{\sqrt{66}} ) leading to ( \theta = 75.7^\circ ) or 1.32 radians.
Step 3
Find an equation of the line of intersection of \( \Pi_1 \) and \( \Pi_2 \)
Answer
To find the line of intersection, we can solve the system of equations representing the two planes. The equations are:
- ( x + 3y - z = 12 ) (from ( \Pi_1 ))\n2. ( 2x - y + z = 10 ) (from ( \Pi_2 ))\n
We solve these simultaneously to express the intersection line in the parametric form. Choose a convenient value (e.g., letting ( z = t )):
By solving for ( x ) and ( y ), we can obtain the general form of the line's equation:
[ \mathbf{r} = (6, 2, 0) + \lambda(-2, -3, -7) ]
where we can isolate coefficients that determine the vector direction.