Use the Question 16 Writing Booklet (a) (i) The point P(x, y, z) lies on the sphere of radius 1 centred at the origin O. Using the position vector of P, $ar{OP} = xar{i} + yar{j} + zar{k}$, and the triangle inequality, or otherwise, show that $|x| + |y| + |z| ightarrow 1$. (ii) Given the vectors $ar{a} = egin{pmatrix} a_1 \ a_2 \ a_3 \\ ar{b} = egin{pmatrix} b_1 \ b_2 \ b_3 \\ $ show that $$ |a_1b_1 + a_2b_2 + a_3b_3| \\ ext{ } \\ ext{ } \\ \\ \\ ext{ } \\ \\ ext{ } \\ \\ ext{ } \\ \\ ext{ } \\ \\ ext{ } \\ \\ ext{ } \\ \\ ext{ } \\ |\sqrt{a_1^2 + a_2^2 + a_3^2} \\\ |\sqrt{b_1^2 + b_2^2 + b_3^2}|$$ (iii) As in part (i), the point P(x, y, z) lies on the sphere of radius 1 centred at the origin O. Using part (ii), show that $|x| + |y| + |z| \leq \sqrt{3}$.

SSCE HSC Mathematics Extension 2 3D vectors: Use the Question 16 Writing Book

Use the Question 16 Writing Booklet (a) (i) The point P(x, y, z) lies on the sphere of radius 1 centred at the origin O - HSC - SSCE Mathematics Extension 2 - Question 16 - 2021 - Paper 1

Question 16

Use the Question 16 Writing Booklet (a) (i) The point P(x, y, z) lies on the sphere of radius 1 centred at the origin O. Using the position vector of P, $ar{OP} =... show full transcript

Worked Solution & Example Answer:Use the Question 16 Writing Booklet (a) (i) The point P(x, y, z) lies on the sphere of radius 1 centred at the origin O - HSC - SSCE Mathematics Extension 2 - Question 16 - 2021 - Paper 1

Step 1

(i) Using the position vector of P, $ar{OP} = xar{i} + yar{j} + zar{k}$, and the triangle inequality, or otherwise, show that $|x| + |y| + |z| ightarrow 1$.

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Answer

To show that the point P lies on the sphere of radius 1, we start with the equation defining the sphere:

$|x|^2 + |y|^2 + |z|^2 = 1.$
By applying the triangle inequality, we know that:

$|x| + |y| + |z| \leq \sqrt{(|x| + |y| + |z|)^2}.$
Thus, we can write:

$|x| + |y| + |z| \leq \sqrt{|x|^2 + |y|^2 + |z|^2 + 2(|xy| + |yz| + |zx|)}.$ Since the maximum value of $|xy| + |yz| + |zx|$ occurs when $|x|, |y|, |z|$ all equal 1 at most, we conclude: $|x| + |y| + |z| \leq 1.$

Step 2

(ii) Given the vectors $ar{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ \end{pmatrix} \text{ and } \bar{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ \end{pmatrix}$, show that $|a_1b_1 + a_2b_2 + a_3b_3| \leq \sqrt{a_1^2 + a_2^2 + a_3^2} \cdot \sqrt{b_1^2 + b_2^2 + b_3^2}$.

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Answer

To prove this inequality, we apply the Cauchy-Schwarz inequality:

$|a_1b_1 + a_2b_2 + a_3b_3| \leq \sqrt{(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2)}.$
This establishes that the inner product of the two vectors is bounded by the product of their magnitudes.

Step 3

(iii) Using part (ii), show that $|x| + |y| + |z| \leq \sqrt{3}$.

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Answer

From part (ii), we know:

$|x| + |y| + |z| \leq \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}.$
This completes the proof that the sum of the absolute values of the coordinates is at most the square root of 3 when the point P lies on the sphere of radius 1.

Use the Question 16 Writing Booklet (a) (i) The point P(x, y, z) lies on the sphere of radius 1 centred at the origin O - HSC - SSCE Mathematics Extension 2 - Question 16 - 2021 - Paper 1

Worked Solution & Example Answer:Use the Question 16 Writing Booklet (a) (i) The point P(x, y, z) lies on the sphere of radius 1 centred at the origin O - HSC - SSCE Mathematics Extension 2 - Question 16 - 2021 - Paper 1

(i) Using the position vector of P, $ar{OP} = xar{i} + yar{j} + zar{k}$, and the triangle inequality, or otherwise, show that $|x| + |y| + |z| ightarrow 1$.

(ii) Given the vectors $ar{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ \end{pmatrix} \text{ and } \bar{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ \end{pmatrix}$, show that $|a_1b_1 + a_2b_2 + a_3b_3| \leq \sqrt{a_1^2 + a_2^2 + a_3^2} \cdot \sqrt{b_1^2 + b_2^2 + b_3^2}$.

(iii) Using part (ii), show that $|x| + |y| + |z| \leq \sqrt{3}$.

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(i) Using the position vector of P, $ar{OP} = xar{i} + yar{j} + zar{k}$, and the triangle inequality, or otherwise, show that $|x| + |y| + |z| ightarrow 1$.

(ii) Given the vectors $ar{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \\ \end{pmatrix} \text{ and } \bar{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ \end{pmatrix}$, show that $|a_1b_1 + a_2b_2 + a_3b_3| \leq \sqrt{a_1^2 + a_2^2 + a_3^2} \cdot \sqrt{b_1^2 + b_2^2 + b_3^2}$.