(a) (i) The point P(x, y, z) lies on the sphere of radius 1 centered at the origin O. Using the position vector of P, $ar{OP} = xi + yj + zk$, and the triangle inequality, or otherwise, show that $|x| + |y| + |z| geq 1$. (ii) Given the vectors $ extbf{a} = \begin{bmatrix} a_1 \ a_2 \ a_3 \ \end{bmatrix}$ and $ extbf{b} = \begin{bmatrix} b_1 \ b_2 \ b_3 \ \end{bmatrix}$, show that $$|\textbf{a}^T \textbf{b}| \leq \sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}.$$ (iii) As in part (ii), the point P(x, y, z) lies on the sphere of radius 1 centered at the origin O. Using part (ii), or otherwise, show that $|x| + |y| + |z| \leq \sqrt{3}$.

SSCE HSC Mathematics Extension 2 3D space: (a) (i) The point P(x, y, z) lie

(a) (i) The point P(x, y, z) lies on the sphere of radius 1 centered 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 centered at the origin O. Using the position vector of P, $ar{OP} = xi + yj + zk$, and the triangle ine... show full transcript

Worked Solution & Example Answer:(a) (i) The point P(x, y, z) lies on the sphere of radius 1 centered 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 $|x| + |y| + |z| geq 1$.

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Answer

To show that $|x| + |y| + |z| geq 1$ , we consider the position vector $ar{OP} = xi + yj + zk$ . The point P lies on the sphere defined by the equation:

$x^2 + y^2 + z^2 = 1.$

By applying the triangle inequality, we have:

$|x| + |y| + |z| \geq |x + y + z|.$

Using the Cauchy-Schwarz inequality:

$(|x| + |y| + |z|)^2 \leq (1 + 1 + 1)(x^2 + y^2 + z^2).$

This simplifies to:

$(|x| + |y| + |z|)^2 \leq 3 \cdot 1 = 3,$

Thus, taking the square root:

$|x| + |y| + |z| \geq 1.$

So, we can conclude that $|x| + |y| + |z| geq 1$ .

Step 2

(ii) Given the vectors a and b, show that $|\textbf{a}^T \textbf{b}| \leq \sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}$.

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Answer

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

$|\textbf{a}^T \textbf{b}| = |a_1b_1 + a_2b_2 + a_3b_3|\leq \sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}.$

This holds true by the definition of the dot product, which adheres to the Cauchy-Schwarz equality condition. This completes the proof.

Step 3

(iii) Show that $|x| + |y| + |z| \leq \sqrt{3}$.

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Answer

In part (ii), we established that:

$|\textbf{a}^T \textbf{b}| \leq \sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}.$

Now, applying this to our case where $x, y, z$ correspond to $extbf{a}$ and $extbf{b}$ respectively, we have:

$|x| + |y| + |z| \leq \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}.$

Thus, we conclude that $|x| + |y| + |z| \leq \sqrt{3}$ .

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

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

(ii) Given the vectors a and b, show that $|\textbf{a}^T \textbf{b}| \leq \sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}$.

(iii) Show that $|x| + |y| + |z| \leq \sqrt{3}$.

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