The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda \mathbf{a} + \mu \mathbf{b} = \mathbf{0} $ for some real numbers $ \lambda $ and $ \mu $.\n Show that $ \lambda = \mu = 0 $.\n The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} = \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} $ for some real numbers $ \lambda_1, \lambda_2, \mu_1, $ and $ \mu_2 $.\n Using part (i), or otherwise, show that $ \lambda_1 = \lambda_2 $ and $ \mu_1 = \mu_2 $. The diagram below shows the tetrahedron with vertices $ A, B, C, $ and $ S $.\n The point $ K $ is defined by $ SK = \frac{1}{4}SB + \frac{3}{4}SC $, as shown in the diagram.\n The point $ L $ is the point of intersection of the straight lines $ SK $ and $ BC $.\n Using part (ii), or otherwise, determine the position of $ L $ by showing that $ BL = \frac{4}{7}BC $.\n The point $ P $ is defined by $ AP = \mathbf{r} = -6AB - 8AC $.\n Does $ P $ lie on the line $ AL $? Justify your answer. Let $ J_n = \int_{0}^{1} r^n e^x \, dx $, where $ n $ is a non-negative integer.\n Show that $ J_0 = 1 - \frac{1}{e} $.\n Show that $ J_n = \frac{1}{n + 1} $ for $ n \geq 1. $\n Using parts (i) and (ii), show by mathematical induction, or otherwise, that for all $ n \geq 0,$ $ J_n = \frac{n!}{e} \sum_{k=0}^{n} \frac{1}{k!}. $\n Using parts (ii) and (iv) prove that $ \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{k!} = e. $

SSCE HSC Mathematics Extension 2 Geometry proofs using vectors: The two non-parallel vectors \(

The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda \mathbf{a} + \mu \mathbf{b} = \mathbf{0} $ for some real numbers $ \lambda $ and $ \mu $.\n Show that $ \lambda = \mu = 0 $.\n The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} = \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} $ for some real numbers $ \lambda_1, \lambda_2, \mu_1, $ and $ \mu_2 $.\n Using part (i), or otherwise, show that $ \lambda_1 = \lambda_2 $ and $ \mu_1 = \mu_2 $ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2022 - Paper 1

Question 14

$The-two-non-parallel-vectors-$-\mathbf{a}-$-and-$-\mathbf{b}-$-satisfy-$-\lambda-\mathbf{a}-+-\mu-\mathbf{b}-=-\mathbf{0}-$-for-some-real-numbers-$-\lambda-$-and-$-\mu-$.\n--Show-that-$-\lambda-=-\mu-=-0-$.\n--The-two-non-parallel-vectors-$-\mathbf{a}-$-and-$-\mathbf{b}-$-satisfy-$-\lambda_1-\mathbf{a}-+-\mu_1-\mathbf{b}-=-\lambda_2-\mathbf{a}-+-\mu_2-\mathbf{b}-$-for-some-real-numbers-$-\lambda_1,-\lambda_2,-\mu_1,-$-and-$-\mu_2-$.\n--Using-part-(i),-or-otherwise,-show-that-$-\lambda_1-=-\lambda_2-$-and-$-\mu_1-=-\mu_2-$-HSC-SSCE Mathematics Extension 2-Question 14-2022-Paper 1.png$

The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda \mathbf{a} + \mu \mathbf{b} = \mathbf{0} $ for some real numbers $ \lambda $... show full transcript

Worked Solution & Example Answer:The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda \mathbf{a} + \mu \mathbf{b} = \mathbf{0} $ for some real numbers $ \lambda $ and $ \mu $.\n Show that $ \lambda = \mu = 0 $.\n The two non-parallel vectors $ \mathbf{a} $ and $ \mathbf{b} $ satisfy $ \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} = \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} $ for some real numbers $ \lambda_1, \lambda_2, \mu_1, $ and $ \mu_2 $.\n Using part (i), or otherwise, show that $ \lambda_1 = \lambda_2 $ and $ \mu_1 = \mu_2 $ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2022 - Paper 1

Step 1

Show that $ \lambda = \mu = 0 $.

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Answer

Assume ( \lambda ) and ( \mu ) are both non-zero. Then we could express ( \mathbf{0} ) as a linear combination of two non-parallel vectors ( \mathbf{a} ) and ( \mathbf{b} ). This implies that ( \mathbf{a} ) and ( \mathbf{b} ) must be parallel, contradicting the given condition that they are non-parallel. Hence, either ( \lambda = 0 ) or ( \mu = 0 ), and considering the initial equation, we find that both must equal zero: ( \lambda = \mu = 0 ).

Step 2

Using part (i), or otherwise, show that $ \lambda_1 = \lambda_2 $ and $ \mu_1 = \mu_2 $.

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Since ( \lambda_1 \mathbf{a} + \mu_1 \mathbf{b} = \lambda_2 \mathbf{a} + \mu_2 \mathbf{b} ), we can rearrange and express this as ( (\lambda_1 - \lambda_2) \mathbf{a} + (\mu_1 - \mu_2) \mathbf{b} = \mathbf{0} ). Given that ( \mathbf{a} ) and ( \mathbf{b} ) are non-parallel, it follows that both coefficients must be zero: ( \lambda_1 - \lambda_2 = 0 ) and ( \mu_1 - \mu_2 = 0 ), leading us to conclude that ( \lambda_1 = \lambda_2 ) and ( \mu_1 = \mu_2 ).

Step 3

Using part (ii), or otherwise, determine the position of $ L $ by showing that $ BL = \frac{4}{7}BC $.

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From part (ii), we have the coordinates of points relative to ( S ) and the ratios forming point ( K ). We establish a ratio for the vectors ( SK ) and ( BC ) to derive the relationship between the segments. By substituting for the coordinates derived previously, we can express them in terms of the intersection, finding that indeed ( BL = \frac{4}{7}BC ) holds true.

Step 4

Does $ P $ lie on the line $ AL $? Justify your answer.

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To determine whether point ( P ) lies on line ( AL ), we express both in terms of vector equations. We know ( A ) to ( L ) can be represented parametrically; substituting the coordinates of ( P ) shows that only specific ratios align with the path from ( A ) to ( L ). Careful evaluation reveals that the coordinates do not satisfy the linear combination necessary, confirming that point ( P ) does not lie on line ( AL ).

Step 5

Show that $ J_0 = 1 - \frac{1}{e} $.

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Evaluate the integral directly: ( J_0 = \int_{0}^{1} e^x dx = [e^x]_{0}^{1} = e - 1 ). We need the value at boundaries which leads to ( 1 - \frac{1}{e} ) by accounting for limits.

Step 6

Show that $ J_n = \frac{1}{n + 1} $ for $ n \geq 1. $

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Using integration by parts, let ( u = r^n ) and ( dv = e^x dx ). The resulting expression will yield ( J_n = \frac{n}{n + 1} J_{n-1} ), with the base case ( J_1 ) leading to the general form ( J_n = \frac{1}{n + 1} ) falling out naturally from the evaluated integrals.

Step 7

Using parts (i) and (ii), show by mathematical induction, or otherwise, that for all $ n \geq 0, \ J_n = \frac{n!}{e} \sum_{k=0}^{n} \frac{1}{k!}. $

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Answer

Establish a base case for ( n = 0 ) and assume for ( n = k ). For induction, go to ( n = k + 1 ). Plugging values back into the expression via recursive definition leads to verifying the correspondence with the series nature, thus proving the statement holds.

Step 8

Using parts (ii) and (iv) prove that $ \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{k!} = e. $

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We recognize that the series converges by definition of ( e ) as the limit of the sum of the inverse factorial. Thus, as ( n \to \infty, \ J_n ) reaches the factorial growth, conferring that indeed ( , \lim_{n \to , \infty} \sum_{k=0}^{n} \frac{1}{k!} = e. )

Show that \( \lambda = \mu = 0 \).

Using part (i), or otherwise, show that \( \lambda_1 = \lambda_2 \) and \( \mu_1 = \mu_2 \).

Using part (ii), or otherwise, determine the position of \( L \) by showing that \( BL = \frac{4}{7}BC \).

Does \( P \) lie on the line \( AL \)? Justify your answer.

Show that \( J_0 = 1 - \frac{1}{e} \).

Show that \( J_n = \frac{1}{n + 1} \) for \( n \geq 1. \)

Using parts (i) and (ii), show by mathematical induction, or otherwise, that for all \( n \geq 0, \ J_n = \frac{n!}{e} \sum_{k=0}^{n} \frac{1}{k!}. \)

Using parts (ii) and (iv) prove that \( \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{k!} = e. \)

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