HSC – unpacking questions
Support for the delivery of mathematics subjects in the lead-up to the HSC examinations.
Intersection of line segments
Watch the video determining the intersection of line segments, answering question 10 from NESA's sample examination (3:37).
Jackie Blue
This is the HSC hub Mathematics curriculum support for the New South Wales Department of Education. My name is Jackie Blue. This video provides a solution to multiple choice question 10 from NESA sample HSC examination for the Extension 2 course. This question looks at vectors. The solution provided in this video demonstrates one way to unpack the question. There may be other methods and we encourage you to discuss alternative solutions with each other.
This question provides four pairs of vector equations of a line within a certain domain. We asked to determine which lines segments intersect at exactly one point. Please press pause now to read the question, and then we'll go through the solution together.
Let's take a look at option A. Each line segment we need to determine the start and end points for vector u. The starting point is given by the position vector one two which can be found when the value of Lambda is equal to zero. The endpoint can be found by calculating the vector equation when the parameter Lambda is equal to one. This gives zero five. Because one plus negative one is zero and two plus three is five. Similarly for Vector V, the starting point is given by the position vector of three one. And calculating the vector equation for Lambda equals one gives four two. It's obvious that option A does not provide a solution to this question.
Moving on to option B. The position vector gives us the starting point. one two And calculating the vector equation when Lambda equals to one gives us the endpoint. zero five In vector v, our starting, point is the position vector five seven. And the end point when Lambda equals one is eight negative two. Once again, it's obvious that they segments do not intersect.
Let's move on to option C For vector u a starting point is given by the position vector zero two The endpoint given by calculating the vector equation when Lambda equals to one. Gives two zero Now for Vector V, the starting point is given by the position vector zero three. And the end point when Lambda equals one gives one zero . We can see here that the lines intersect at exactly one point, so the correct answer is C.
This is the HSC hub for the New South Wales Department of Education.
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Using the scalar product
Watch the video using the scalar product, answering question 15b of NESA's sample examination (3:33).
Daniel Proctor
This is the HSC hub Mathematics curriculum support for the New South Wales Department of Education. My name is Daniel Proctor. This video provides a solution to question fifteen b from the sample examination provided by the New South Wales Education Standards Authority for the Mathematics Extension 2 course. This question looks at vectors.
The solution provided in this video demonstrates one way to unpack a question there may be other methods, and we encourage you to discuss any alternative solutions with your teacher. This question shows a position vector away in 3 dimensions. Click pause now to read the question and consider some of the key pieces of information in the question.
The first point to note is the allocation of four marks to this question. This indicates there will be multiple stages that need to be communicated clearly. Inspecting the information in the question shows vectors are represented as a column vector. And unit factors IJ&K. The question provides some direction to the method to be used. In this case the dot product.
Lastly, supplement the information in the diagram with the information in the question. It is not easy to read the angles Alpha, Beta and Gamma from the diagram and it only becomes clear when you read it in conjunction with the text. This question asks us to prove the result cos squared Alpha plus cos squared beta plus cos squared gamma is equal to one
To start will define the vector away using unit factors. Applying the geometric form of the dot product to A and the unit vector, I give this result and linking this to the algebraic form of the dot product gives this. Which simplifies, to this result. Attempting to determine this result, or one mark out of a possible four. Similar results can be determined using both forms of the dot product of OA with the unit vectors J&K. And determining all three equations would gain 2 marks.
Drawing links to the result that needs to be proved directs us to square both sides of each equation to form expressions with Cos squared terms. Adding the left hand sides, and the right of these three equations collects the three cos squared terms into one equation, like the one to be proved. Determining this equation would gain 3 marks. The final mark can be achieved by realizing that the magnitude of the vector OA is equal to the square root of the sum of the squares AB&C. Substituting this result into the equation and simplifying generates the result needed.
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