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Find a cartesian equation of the plane \( \Pi \) containing the lines \( \mathbf{r} = 3\mathbf{i} + \mathbf{k} + s(2\mathbf{i} + \mathbf{j} - \mathbf{k}) \) and \( \mathbf{r} = -3\mathbf{i} - 7\mathbf{j} + 10\mathbf{k} + t(-\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) \) - CIE - A-Level Further Maths - Question 9 - 2011 - Paper 1
Question 9
Find a cartesian equation of the plane \( \Pi \) containing the lines \( \mathbf{r} = 3\mathbf{i} + \mathbf{k} + s(2\mathbf{i} + \mathbf{j} - \mathbf{k}) \) and \( ... show full transcript
Worked Solution & Example Answer:Find a cartesian equation of the plane \( \Pi \) containing the lines \( \mathbf{r} = 3\mathbf{i} + \mathbf{k} + s(2\mathbf{i} + \mathbf{j} - \mathbf{k}) \) and \( \mathbf{r} = -3\mathbf{i} - 7\mathbf{j} + 10\mathbf{k} + t(-\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) \) - CIE - A-Level Further Maths - Question 9 - 2011 - Paper 1
Step 1
Find the cartesian equation of the plane \( \Pi \)
Answer
To find the equation of the plane containing the lines, we first determine the direction vectors of the given lines. The direction vector of the first line is ( \mathbf{d_1} = 2\mathbf{i} + \mathbf{j} - \mathbf{k} ), and the second line has the direction vector ( \mathbf{d_2} = -\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} ).
Next, we find the normal vector of the plane by calculating the cross product of the direction vectors:
\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ -1 & -3 & 4 \end{vmatrix}$$ Calculating this determinant gives \( \mathbf{n} = \mathbf{i}(-3 + 4) - \mathbf{j}(-8 + 1) + \mathbf{k}(-6 - 1) = \mathbf{i} + 7\mathbf{j} - 7\mathbf{k}. \) Therefore, the normal vector is \( \mathbf{n} = 2\mathbf{i} + 1\mathbf{j} - 3\mathbf{k}. \) The equation of the plane can be expressed as: $$2(x - x_0) + 1(y - y_0) - 3(z - z_0) = 0$$ Substituting a point from either line (like \( \mathbf{r} = -3\mathbf{i} - 7\mathbf{j} + 10\mathbf{k} \)) leads to the final equation.Step 2
(i) the position vector of the point where \( \ell \) meets \( \Pi \)
Answer
To find the intersection of line ( \ell ) and plane ( \Pi ), substitute the parameterized form of line ( \ell ):
( \mathbf{r} = (6 + 2\lambda)\mathbf{i} + (-2 + \lambda) extbf{j} + (1 - 4\lambda)\mathbf{k} ) into the plane equation we found earlier.
Solving for ( \lambda ) will give us the intersection point's position vector:
Thus, computing will lead us to find the vector at ( \lambda = 1 ), yielding ( 4\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}. )
Step 3
(ii) the perpendicular distance from \( P \) to \( \Pi \)
Answer
The distance ( D ) from point ( P ) to the plane can be calculated using the formula:
where ( \mathbf{r}_0 ) is position vector of point ( P ) and ( \mathbf{n} ) is the normal vector. Substituting the values gives the perpendicular distance.
Step 4
(iii) the acute angle between \( \ell \) and \( \Pi \)
Answer
To find the acute angle ( \theta ) between line ( \ell ) and plane ( \Pi ):
The angle can be found using the formula:
Where ( \mathbf{d} ) is the direction vector of line ( \ell ). Upon calculation, we will find ( \theta ) to be approximately ( 0.236 \text{ radians} ).