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Use the Question 16 Writing Booklet (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 (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, \( \overrighta... show full transcript
Worked Solution & Example Answer:Use the Question 16 Writing Booklet (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, \( \overrightarrow{OP} = xi + yj + zk \), and the triangle inequality, or otherwise, show that \( |x| + |y| + |z| \geq 1 \).
Answer
To show that ( |x| + |y| + |z| \geq 1 ), we start by noting that the point P lies on the sphere with equation:
Using the triangle inequality:
[ |x| + |y| + |z| \geq \sqrt{x^2 + y^2 + z^2} = \sqrt{1} = 1. ]
Thus, it follows that ( |x| + |y| + |z| \geq 1 ).
Step 2
(ii) Given the vectors \( \mathbf{a} = \begin{pmatrix} a_1 \ a_2 \ a_3 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_1 \ b_2 \ b_3 \end{pmatrix} \), show that \( |\mathbf{a}^T \mathbf{b}| \leq \sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}. \)
Answer
Using the Cauchy-Schwarz inequality, we can write:
[ |\mathbf{a}^T \mathbf{b}| = |\begin{pmatrix} a_1 & a_2 & a_3 \end{pmatrix} \begin{pmatrix} b_1 \ b_2 \ b_3 \end{pmatrix}| \leq \sqrt{a_1^2 + a_2^2 + a_3^2} \cdot \sqrt{b_1^2 + b_2^2 + b_3^2}. ]
Thus, we have shown the desired result.
Step 3
(iii) Using part (ii), or otherwise, show that \( |x| + |y| + |z| \leq \sqrt{3}. \)
Answer
From part (ii), we know that for the point P on the sphere:
[ |x| + |y| + |z| \leq \sqrt{(|x|^2 + |y|^2 + |z|^2)(1 + 1 + 1)}. ]
However, since ( x^2 + y^2 + z^2 = 1 ), we can substitute in:
[ |x| + |y| + |z| \leq \sqrt{1 \cdot 3} = \sqrt{3}. ]
Therefore, we conclude that ( |x| + |y| + |z| \leq \sqrt{3}. )