Home / CIE A level -Pure Mathematics 3 : Topic : 3.7 Vectors-standard notations for vectors, : Exam Style Questions Paper 3

CIE A level -Pure Mathematics 3 : Topic : 3.7 Vectors-standard notations for vectors, : Exam Style Questions Paper 3

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 

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