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NEW Education Standards Authority 2018 HIGHER SCHOOL CERTIFICATE EXAMINATION Mathematics Extension 1 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2018 - Paper 1
Question 1
NEW Education Standards Authority 2018 HIGHER SCHOOL CERTIFICATE EXAMINATION Mathematics Extension 1
Worked Solution & Example Answer:NEW Education Standards Authority 2018 HIGHER SCHOOL CERTIFICATE EXAMINATION Mathematics Extension 1 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2018 - Paper 1
Step 1
Part a: State the condition for the product of two vectors to be zero.
Answer
For the product of two vectors \( extbf{a} \) and \( extbf{b} \) to be zero, it is necessary that either one of the vectors is the zero vector or they are orthogonal to each other. Mathematically, this can be stated as:
\[ extbf{a} \cdot extbf{b} = 0 \ ext{ or } \textbf{a} = extbf{0} \ ext{ or } \textbf{b} = extbf{0} ]}
This implies that the angle between the two vectors is 90 degrees if neither vector is the zero vector.
Step 2
Part b: Calculate the cross product of two specific vectors.
Answer
Let \( extbf{a} = egin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} \) and \( extbf{b} = egin{pmatrix} 4 \ 5 \ 6 \end{pmatrix} \).
The cross product \( extbf{a} \times extbf{b} \) is calculated using the determinant of a matrix:
\[ extbf{a} \times extbf{b} = egin{vmatrix} extbf{i} & extbf{j} & extbf{k} \ 1 & 2 & 3 \ 4 & 5 & 6 \end{vmatrix} \]
Calculating this determinant gives:
\[ extbf{a} \times extbf{b} = egin{pmatrix} 2 \cdot 6 - 3 \cdot 5 \negative \newline 3 \cdot 4 - 1 \cdot 6 \newline 1 \cdot 5 - 2 \cdot 4 \end{pmatrix} = egin{pmatrix} -3 \newline -6 \newline -3 \end{pmatrix} \]
Thus, the cross product of \( extbf{a} \) and \( extbf{b} \) is \(-3 \textbf{i} - 6 \textbf{j} - 3 \textbf{k}.")]}]} ```json {