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Question
Consider the lines \({L_1}\) and \({L_2}\) with equations \({L_1}\) : \(\boldsymbol{r}=\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ – 1\end{array} \right)\) and \({L_2}\) : \(\boldsymbol{r} = \left( \begin{array}{c}1\\1\\ – 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)\).
The lines intersect at point \(\rm{P}\).
Find the coordinates of \({\text{P}}\).
Show that the lines are perpendicular.
The point \({\text{Q}}(7, 5, 3)\) lies on \({L_1}\). The point \({\text{R}}\) is the reflection of \({\text{Q}}\) in the line \({L_2}\).
Find the coordinates of \({\text{R}}\).
Answer/Explanation
Markscheme
appropriate approach (M1)
eg \(\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ – 1\end{array} \right) = \left( \begin{array}{c}1\\1\\ – 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)\), \({L_1} = {L_2}\)
any two correct equations A1A1
eg \(11 + 4s = 1 + 2t,{\text{ }}8 + 3s = 1 + t,{\text{ }}2 – s = – 7 + 11t\)
attempt to solve system of equations (M1)
eg \(10 + 4s = 2(7 + 3s), \left\{ {\begin{array}{*{20}{c}} {4s – 2t = – 10} \\ {3s – t = – 7} \end{array}} \right.\)
one correct parameter A1
eg \(s = – 2,{\text{ }}t = 1\)
\({\text{P}}(3, 2, 4)\) (accept position vector) A1 N3
[6 marks]
choosing correct direction vectors for \({L_1}\) and \({L_2}\) (A1)(A1)
eg \(\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right),\left( {\begin{array}{*{20}{c}} 2 \\ 1 \\ {11} \end{array}} \right)\) (or any scalar multiple)
evidence of scalar product (with any vectors) (M1)
eg \(a \cdot b\), \(\left( \begin{array}{c}4\\3\\ – 1\end{array} \right) \bullet \left( \begin{array}{c}2\\1\\11\end{array} \right)\)
correct substitution A1
eg \(4(2) + 3(1) + ( – 1)(11),{\text{ }}8 + 3 – 11\)
calculating \(a \cdot b = 0\) A1
Note: Do not award the final A1 without evidence of calculation.
vectors are perpendicular AG N0
[5 marks]
Note: Candidates may take different approaches, which do not necessarily involve vectors.
In particular, most of the working could be done on a diagram. Award marks in line with the markscheme.
METHOD 1
attempt to find \(\overrightarrow {{\text{QP}}} \) or \(\overrightarrow {{\text{PQ}}} \) (M1)
correct working (may be seen on diagram) A1
eg \(\overrightarrow {{\text{QP}}} \) = \(\left( \begin{array}{c} – 4\\ – 3\\1\end{array} \right)\), \(\overrightarrow {{\text{PQ}}} \) = \(\left( \begin{array}{c}7\\5\\3\end{array} \right) – \left( \begin{array}{c}3\\2\\4\end{array} \right)\)
recognizing \({\text{R}}\) is on \({L_1}\) (seen anywhere) (R1)
eg on diagram
\({\text{Q}}\) and \({\text{R}}\) are equidistant from \({\text{P}}\) (seen anywhere) (R1)
eg \(\overrightarrow {{\text{QP}}} = \overrightarrow {{\text{PR}}} \), marked on diagram
correct working (A1)
eg \(\left( \begin{array}{c}3\\2\\4\end{array} \right) – \left( \begin{array}{c}7\\5\\3\end{array} \right) = \left( \begin{array}{c}x\\y\\z\end{array} \right) – \left( \begin{array}{c}3\\2\\4\end{array} \right),\left( \begin{array}{c} – 4\\ – 3\\1\end{array} \right) + \left( \begin{array}{c}3\\2\\4\end{array} \right)\)
\({\text{R}}(–1, –1, 5)\) (accept position vector) A1 N3
METHOD 2
recognizing \({\text{R}}\) is on \({L_1}\) (seen anywhere) (R1)
eg on diagram
\({\text{Q}}\) and \({\text{R}}\) are equidistant from \({\text{P}}\) (seen anywhere) (R1)
eg \({\text{P}}\) midpoint of \({\text{QR}}\), marked on diagram
valid approach to find one coordinate of mid-point (M1)
eg \({x_p} = \frac{{{x_Q} + {x_R}}}{2},{\text{ }}2{y_p} = {y_Q} + {y_R},{\text{ }}\frac{1}{2}\left( {{z_Q} + {z_R}} \right)\)
one correct substitution A1
eg \({x_R} = 3 + (3 – 7),{\text{ }}2 = \frac{{5 + {y_R}}}{2},{\text{ }}4 = \frac{1}{2}(z + 3)\)
correct working for one coordinate (A1)
eg \({x_R} = 3 – 4,{\text{ }}4 – 5 = {y_R},{\text{ }}8 = (z + 3)\)
\({\text{R}} (-1, -1, 5)\) (accept position vector) A1 N3
[6 marks]
Question
Two points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.
Let \({\mathop {{\text{PR}}}\limits^ \to }\) = 6i − j + 3k.
Find \(\mathop {{\text{PQ}}}\limits^ \to \).
Find \(\left| {\mathop {{\text{PQ}}}\limits^ \to } \right|\).
Find the angle between PQ and PR.
Find the area of triangle PQR.
Hence or otherwise find the shortest distance from R to the line through P and Q.
Answer/Explanation
Markscheme
valid approach (M1)
eg (7, 4, 9) − (3, 2, 5) A − B
\(\mathop {{\text{PQ}}}\limits^ \to = \) 4i + 2j + 4k \(\left( { = \left( \begin{gathered}
4 \hfill \\
2 \hfill \\
4 \hfill \\
\end{gathered} \right)} \right)\) A1 N2
[2 marks]
correct substitution into magnitude formula (A1)
eg \(\sqrt {{4^2} + {2^2} + {4^2}} \)
\(\left| {\mathop {{\text{PQ}}}\limits^ \to } \right| = 6\) A1 N2
[2 marks]
finding scalar product and magnitudes (A1)(A1)
scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)
magnitude of PR = \(\sqrt {36 + 1 + 9} = \left( {6.782} \right)\)
correct substitution of their values to find cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\) M1
eg cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\,\,{\text{ = }}\frac{{24 – 2 + 12}}{{\left( 6 \right) \times \left( {\sqrt {46} } \right)}},\,\,0.8355\)
0.581746
\({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\) = 0.582 radians or \({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\) = 33.3° A1 N3
[4 marks]
correct substitution (A1)
eg \(\frac{1}{2} \times \left| {\mathop {{\text{PQ}}}\limits^ \to } \right| \times \left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\frac{1}{2} \times 6 \times \sqrt {46} \times \,\,{\text{sin}}\,0.582\)
area is 11.2 (sq. units) A1 N2
[2 marks]
recognizing shortest distance is perpendicular distance from R to line through P and Q (M1)
eg sketch, height of triangle with base \(\left[ {{\text{PQ}}} \right],\,\,\frac{1}{2} \times 6 \times h,\,\,{\text{sin}}\,33.3^\circ = \frac{h}{{\sqrt {46} }}\)
correct working (A1)
eg \(\frac{1}{2} \times 6 \times d = 11.2,\,\,\left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\sqrt {46} \times \,\,{\text{sin}}\,33.3^\circ \)
3.72677
distance = 3.73 (units) A1 N2
[3 marks]