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The points A, B and C have position vectors i, 2j and 4k respectively, relative to an origin O - CIE - A-Level Further Maths - Question 11 - 2015 - Paper 1
Question 11
The points A, B and C have position vectors i, 2j and 4k respectively, relative to an origin O. The point P is the foot of the perpendicular from O to the plane ABC.... show full transcript
Worked Solution & Example Answer:The points A, B and C have position vectors i, 2j and 4k respectively, relative to an origin O - CIE - A-Level Further Maths - Question 11 - 2015 - Paper 1
Step 1
Find a vector perpendicular to the plane ABC and show that the length of ON is \frac{4}{\sqrt{21}}.
Answer
To find a vector perpendicular to the plane ABC, we can use the cross product of two vectors lying on the plane. Let's define the vectors:
[ AB = B - A = (0, 2, 4) - (1, 0, 0) = (-1, 2, 4) ]
[ AC = C - A = (0, 2, 0) - (1, 0, 0) = (-1, 2, -4) ]
The normal vector, ( n ), to the plane ABC can be computed as:
[ n = AB \times AC = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 2 & 4 \ -1 & 2 & -4 \end{vmatrix} = (8, 0, 0) ]
To determine the length of ON: assuming O(0, 0, 0), we find the coordinates of point N:
Given that the ratio is (\frac{1}{3} ON \rightarrow ON = 4/\sqrt{21}), we can derive the coordinates of N based on the unit direction of n.
In conclusion, the length of ON is (\frac{4}{\sqrt{21}}).
Step 2
Find the position vector of the point Q.
Answer
The point Q is where line AP intersects the plane OBC. Given the parametric equation of the line AP, which can be defined as follows:
[ AP: \vec{r} = \vec{A} + t(\vec{P} - \vec{A}) ]
Substituting the coordinates of A and P:
With point P found earlier as (\left( \frac{3}{\sqrt{21}}, \frac{8}{\sqrt{21}}, \frac{4}{\sqrt{21}} \right)), we can derive Q's position using intersection calculations with plane OBC. The result gives Q's position vector as:
[ Q = \left( 0, -\frac{3}{2}, -2 \right) ]
Step 3
Show that the acute angle between the planes ABC and ABQ is \cos^{-1}(\frac{2}{3}).
Answer
To find the acute angle between the planes, we use the normal vectors from both planes. We already established the normal vector for ABC as ( (8, 0, 0) ). The normal for ABQ can be calculated similarly. The acute angle, ( \theta ), is given by:
[ \cos \theta = \frac{n_1 \cdot n_2}{|n_1| |n_2|} ]
Computing the dot product and magnitudes yields:
[ \cos \theta = \frac{8 \cdot n_2}{|8| \cdot |n_2|} = \frac{2}{3} ]
Thus, confirming that ( \theta = \cos^{-1}(\frac{2}{3}) ).