Qurstion
The straight line l has equation r = 4i − j + 2k + , \(\lambda (2i− 3j + 6k)\). The plane p passes through the point( 4, −1, 2 )and is perpendicular to l.
(i) Find the equation of p, giving your answer in the form ax + by + cÏ = d.
(ii) Find the perpendicular distance from the origin to p.
(iii) A second plane q is parallel to p and the perpendicular distance between p and q is 14 units.
Find the possible equations of q.
Answer/Explanation
(i)
(i) Obtain 2x – 3y + 6z for LHS of equation
Obtain 2x – 3y + 6z = 23
(ii) Either Use correct formula to find perpendicular distance
Obtain unsimplified value \(frac{\pm 23}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}\), following answer to (i)
Obtain \(\frac{23}{7}\)or equivalent
OR 1 Use scalar product of (4, –1, 2) and a vector normal to the plane
Use unit normal to plane to obtain\( \pm\frac{(8+3+12){\sqrt{49}}\)
Obtain distance\( \frac{23}{7}\) or equivalent
OR 2 Find parameter intersection of p and \(r=\mu (2i-3j+6k)\)
Obtain \(\mu =\frac{23}{49 }[and\left ( \frac{46}{49},-\frac{69}{49},\frac{138}{49} \right )\)as foot of perpendicular]
Obtain distance \(\frac{23}{7}\) or equivalent
(iii) Either Recognise that plane is 2x – 3y + 6z = k and attempt use of formula for
perpendicular distance to plane at least once
Obtain\( \frac{|23-k|}{7}=14 \) or equivalent
Obtain 2x – 3y + 6z = 121 and 2x – 3y + 6z = –75
OR Recognise that plane is 2x – 3y + 6z = k and attempt to find at least one
point on q using l with λ = ±2
Obtain 2x – 3y + 6z = 121
Obtain 2x – 3y + 6z = –75
Question
The diagram shows three points A, B and C whose position vectors with respect to the origin O are given by \(\underset{OA}{\rightarrow}=\begin{pmatrix}2\\-1\\2\\\end{pmatrix}, \underset{OB}{\rightarrow}=\begin{pmatrix}0\\3\\1\\\end{pmatrix}\) and \(\underset{OC}{\rightarrow}=\begin{pmatrix}3\\0\\4\\\end{pmatrix}\). The point D lies on BC, between B and C, and is such that CD = 2DB.
(i) Find the equation of the plane ABC, giving your answer in the form ax + by + cz = d. [6]
(ii) Find the position vector of D. [1]
(iii) Show that the length of the perpendicular from A to OD is \(\frac{1}{3}\sqrt{\left ( 65 \right )}\).[4]
Answer/Explanation
Ans:
Question
The points P and Q have position vectors, relative to the origin O, given by
\(\underset{OP}{\rightarrow}=7i+7j-5k\) and \(\underset{OQ}{\rightarrow}=5i+j+k.
The mid-point of PQ is the point A. The plane \(\mathbb{I}\) is perpendicular to the line PQ and passes through A.
(i) Find the equation of \(\mathbb{I}\), giving your answer in the form ax + by + cz = d. [4]
(ii) The straight line through P parallel to the x-axis meets \(\mathbb{I}\) at the point B. Find the distance AB, correct to 3 significant figures. [5]
Answer/Explanation
Ans:
(i) State or imply A is (1, 4, –2)
State or imply \(\underset{QP}{\rightarrow}=12i+6j-6k\) or equivalent
Use QP as normal and A as mid-point to find equation of plane
Obtain 12x + 6y – 6z = 48 or equivalent
(ii) Either State equation of PB is r = 7i + 7j – 5k + λi
Set up and solve a relevant equation for λ .
Obtain λ = −9 and hence B is (–2, 7, –5)
Use correct method to find distance between A and B.
Obtain 5.20
Or Obtain 12 for result of scalar product of QP and i or equivalent
Use correct method involving moduli, scalar product and cosine to find angle APB
Obtain 35.26° or equivalent
Use relevant trigonometry to find AB
Obtain 5.20