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Vectors: magnitude, direction and products

Practice section 7 of 9

Questions 37–42 in this collection

Practice questions

Discussion practice · 1 of 6

Solve perpendicularity conditions and compare two area methods

The points AA, BB, CC and DD have position vectors i−2j+5k\mathbf i-2\mathbf j+5\mathbf k, i+3j\mathbf i+3\mathbf j, 10i+j+2k10\mathbf i+\mathbf j+2\mathbf k and −2i+4j+5k-2\mathbf i+4\mathbf j+5\mathbf k respectively, with respect to an origin OO. The point PP on ABAB is such that AP:PB=λ:1−λAP:PB=\lambda:1-\lambda and the point QQ on CDCD is such that CQ:QD=μ:1−μCQ:QD=\mu:1-\mu. Find OP→\overrightarrow{OP} and OQ→\overrightarrow{OQ} in terms of λ\lambda and μ\mu respectively.

Given that PQ→\overrightarrow{PQ} is perpendicular to both AB→\overrightarrow{AB} and CD→\overrightarrow{CD},

(i) show that PQ→=i+2j+2k\overrightarrow{PQ}=\mathbf i+2\mathbf j+2\mathbf k,

(ii) find the area of the triangle ABQABQ.

[2006/I/14(OR)]

CJC · Discussion · Question 11

Solution

Discussion practice · 2 of 6

Connect a cross-product relation to areas and a nonzero linear combination

Relative to the origin OO, three distinct fixed points AA, BB and CC have position vectors a\mathbf a, b\mathbf b and c\mathbf c respectively. It is known that b\mathbf b is a unit vector, ∣a∣=3|\mathbf a|=3, ∣c∣=2|\mathbf c|=2 and the angle AOCAOC is 60∘60^\circ.

(i) State the geometrical interpretation of ∣b⋅c∣|\mathbf b\cdot\mathbf c|. [1]

It is further given that a×2b=b×c\mathbf a\times2\mathbf b=\mathbf b\times\mathbf c.

(ii) Find the ratio of the area of triangle AOBAOB to the area of triangle BOCBOC. [2]

(iii) Show that 2a+c=kb2\mathbf a+\mathbf c=k\mathbf b where k∈Rk\in\mathbb R, k≠0k\ne0.

By considering (2a+c)⋅(2a+c)(2\mathbf a+\mathbf c)\cdot(2\mathbf a+\mathbf c), find the exact values of kk. [6]

[2016/HCI/Promo/7]

CJC · Discussion · Question 12

Solution

Review practice · 3 of 6

Use one displacement for four vector tasks

The position vectors of A, B and C are a=i−j+2k\mathbf a=\mathbf i-\mathbf j+2\mathbf k, b=3i+5j−2k\mathbf b=3\mathbf i+5\mathbf j-2\mathbf k and c=pj+4k\mathbf c=p\mathbf j+4\mathbf k respectively.

(i) Find the unit vector in the direction of AB→\overrightarrow{AB}. [2]

(ii) Find the value of p such that A, B and C are collinear. [3]

(iii) If p=−1p=-1, find the position vector of D such that ABCD is a parallelogram. [2]

(iv) If ∣AC→∣=3|\overrightarrow{AC}|=3, find the possible values of p. [2]

CJC · Review problems · Question 1

Solution

Review practice · 4 of 6

Find a displacement and prove a line intersection ratio

Relatively to an origin O, the position vectors of A, B and P are a\mathbf a, b\mathbf b and 12(a+b)\frac12(\mathbf a+\mathbf b) respectively, where a and b are non-zero and non-parallel vectors. The point Q lies on the line segment OP such that OQ:QP=2:1. The line AQ cuts OB at R.

(i) Find, in terms of a and b, AQ→\overrightarrow{AQ}. [2]

(ii) Show that OR→=12b\overrightarrow{OR}=\frac12\mathbf b. [3]

[2012/MJC/Promo/2]

CJC · Review problems · Question 2

Solution

Review practice · 5 of 6

Recover a dot product, projection and a parameterised area

Referred to the origin O, the points A and B have position vectors a and b respectively such that ∣a∣=5|\mathbf a|=5 and ∣b∣=3|\mathbf b|=3.

(i) It is given that ∣2b−a∣=7|2\mathbf b-\mathbf a|=7. Using a suitable scalar product, show that a⋅b=3\mathbf a\cdot\mathbf b=3. [2]

(ii) Deduce the length of projection of OA→\overrightarrow{OA} onto OB→\overrightarrow{OB}. [1]

The point P lies on AB such that AP:PB=1:k, where k is a positive constant.

(iii) Write down the position vector of P in terms of k,a and b and show that the area of triangle OAP is 36k+1\dfrac{3\sqrt6}{k+1}. [5]

[2022/NYJC/Promo/5]

CJC · Review problems · Question 3

Solution

Review practice · 6 of 6

Find a perpendicular foot on a parameterised line

With respect to O, A,B,C have position vectors a=5i+4j+10k\mathbf a=5\mathbf i+4\mathbf j+10\mathbf k, b=−4i+4j−2k\mathbf b=-4\mathbf i+4\mathbf j-2\mathbf k, c=−5i+9j+5k\mathbf c=-5\mathbf i+9\mathbf j+5\mathbf k.

(i) Find the position vector of D such that ABCD is a parallelogram. [2]

(ii) N divides AB in the ratio λ:(1−λ)\lambda:(1-\lambda). Show that CN→=(10−9λ)i−5j+(5−12λ)k\overrightarrow{CN}=(10-9\lambda)\mathbf i-5\mathbf j+(5-12\lambda)\mathbf k. Hence find the position vector of N given CN is perpendicular to AB. [4]

(iii) P has position vector (−1+3t)i+4j+(2+4t)k(-1+3t)\mathbf i+4\mathbf j+(2+4t)\mathbf k, where t∈Rt\in\mathbb R. Show that P is collinear with A and B for all real t. [3]

[2012/SRJC/Promo/11]

CJC · Review problems · Question 4

Solution

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