Consider the points \(A(1, 0, 2)\), \(B(3, 1, 4)\) and \(C(0, 5, 3)\).
(a) Find the vector \(\vec{AB}\) and the vector \(\vec{AC}\). [2]
(b) Find the vector product \(\vec{AB} \times \vec{AC}\). [3]
(c) Hence, find the area of triangle \(ABC\). [2]
(d) Find the Cartesian equation of the plane \(\Pi\) containing the points A, B and C. [3]
The line \(L\) has vector equation \(\mathbf{r} = (2, -1, 0) + t(1, -2, 1)\), where \(t \in \mathbb{R}\).
(e) Show that the line \(L\) intersects the plane \(\Pi\). [4]
(f) Find the coordinates of the point of intersection \(P\) of the line \(L\) and the plane \(\Pi\). [3]
(g) Find the angle between the line \(L\) and the plane \(\Pi\). [2]
Question
Find the vector \(\vec{AB}\) and the vector \(\vec{AC}\). [2]
Hint
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Hint
Subtract coordinates: \(\vec{AB}=B-A\) and \(\vec{AC}=C-A\).
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\(\vec{AB}=(2,1,2)\) and \(\vec{AC}=(-1,5,1)\)
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Find the vector product \(\vec{AB} \times \vec{AC}\). [3]
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Use the determinant formula for the cross product.
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\(\vec{AB} \times \vec{AC}=(-9,-4,11)\)
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Hence, find the area of triangle \(ABC\). [2]
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Area is half of the magnitude of the cross product.
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Area = \(\frac{1}{2}\sqrt{218}\)
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Find the Cartesian equation of the plane \(\Pi\) containing points A, B, and C. [3]
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Use the normal vector from the cross product with point A to derive the equation.
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Plane equation: \(-9x – 4y + 11z = 13\)
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Show that the line \(L\) intersects the plane \(\Pi\). [4]
Hint
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Hint
Substitute \(r=(2+t, -1-2t, t)\) into the plane equation and solve for t.
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Substitution yields a unique \(t=2.7\), proving that L intersects \(\Pi\).
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Find the coordinates of the point of intersection \(P\) of the line \(L\) and the plane \(\Pi\). [3]
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Use the found value \(t=2.7\) in the line equations to determine P.
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Point \(P = (4.7, -6.4, 2.7)\)
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Find the angle between the line \(L\) and the plane \(\Pi\). [2]
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Hint
Find the angle between the line’s direction vector and the plane’s normal vector, then use the complement.
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Angle between line and plane is approximately \(16.0°\)
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