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Vectors and matrices — Question 7

QCAA 2022, Paper 2 · 1 mark

Q7 · 2022 · Technology-activeSimple familiar1 mark

QUESTION 7

Given

\mathbf a=(3n+2)\hat{\mathbf i}+2\hat{\mathbf j}

,

\mathbf b=(n-2)\hat{\mathbf j}

and

\mathbf a\times\mathbf b=(1-2n)\hat{\mathbf k}

, the possible values of

n

are

(A)

-5

and

\dfrac13

(B)

-1

and

\dfrac53

(C)

1

and

-\dfrac53

(D)

5

and

-\dfrac13

ABCD

Question linkOriginal paper

WORKED SOLUTION

Answer B

1 mark

ANSWER

B

Worked solution

The

k

-component of

\mathbf a\times\mathbf b

is

(3n+2)(n-2)

. Hence

(3n+2)(n-2)=1-2n

, so

3n^2-2n-5=0=(3n-5)(n+1)

. Thus

n=-1

or

n=5/3

.

Selects option B.

[1 mark]

Worked explanation by QCE Vault; correct option verified against the QCAA answer key.

View the full QCAA marking guide

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