(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, $ar{OP} = xi + yj + zk$, and the triangle inequality, or otherwise, show that $|x| + |y| + |z| ullet 1.$ (ii) Given the vectors a = \begin{pmatrix} a_1 \ a_2 \ a_3 \ \end{pmatrix} and b = \begin{pmatrix} b_1 \ b_2 \ b_3 \ \end{pmatrix}, \ show that |a^{ ext{T}}b| \leq \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), or otherwise, show that |x| + |y| + |z| \leq \sqrt{3}.

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(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

(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, $ar{OP} = xi + yj + zk$, and the triangle in... show full transcript

Worked Solution & Example Answer:(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, show that $|x| + |y| + |z| ullet 1$.

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Answer

To show that $|x| + |y| + |z| ullet 1$ , we start by noting that the point P lies on the sphere of radius 1, which is mathematically expressed as:

$|OP| = ext{radius} = 1.$

The distance from the origin O to point P can be calculated as:

$|OP| = \sqrt{x^2 + y^2 + z^2} = 1.$

We can apply the triangle inequality, which states that for any real numbers a, b, and c:

$|a + b + c| \leq |a| + |b| + |c|.$

By applying this to the coordinates, we have:

$|x| + |y| + |z| \geq \sqrt{x^2 + y^2 + z^2} = 1.$

This leads us to conclude that $|x| + |y| + |z| \geq 1$ .

Step 2

(ii) Given the vectors, show that $|a^{ ext{T}}b| \leq \sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}$.

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Answer

Given the vectors:

$a = \begin{pmatrix} a_1 \ a_2 \ a_3 \ \end{pmatrix}, \ b = \begin{pmatrix} b_1 \ b_2 \ b_3 \ \end{pmatrix}$ ,

we want to show that:

$|a^{ ext{T}}b| = |a_1b_1 + a_2b_2 + a_3b_3|.$

Applying the Cauchy-Schwarz inequality, we find:

$|a^{ ext{T}}b| \leq ||a||_2 ||b||_2,$

where:

$||a||_2 = \sqrt{a_1^2 + a_2^2 + a_3^2}, \ ||b||_2 = \sqrt{b_1^2 + b_2^2 + b_3^2}.$

Thus, substituting these back in, we get:

$|a^{ ext{T}}b| \leq \sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}.$

Step 3

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

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Answer

Given that point P lies on the sphere of radius 1, we can express:

$|x| + |y| + |z| \leq \sqrt{(|x| + |y| + |z|)^2}.$

From part (ii), applying the Cauchy-Schwarz inequality as before:

$|(1, 1, 1)^{ ext{T}}(x, y, z)| \leq ||(1, 1, 1)||_2 ||(x, y, z)||_2.$

Calculating these norms:

$||(1, 1, 1)||_2 = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

and, since the radius is 1, we have:

$||(x, y, z)||_2 = |x| + |y| + |z| \leq 1.$

Thus, we deduce:

$|x| + |y| + |z| \leq \sqrt{3}.$

(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:(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, show that $|x| + |y| + |z| ullet 1$.

(ii) Given the vectors, show that $|a^{ ext{T}}b| \leq \sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}$.

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

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(i) Using the position vector of P, show that $|x| + |y| + |z| ullet 1$.