Find the value of t for which BQM is a straight line

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TSR360
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#1
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#1
Question (part c): http://prntscr.com/s7jyig
My w/o: http://prntscr.com/s7k0hx

I don't see how I can find the value of t unless I know the value of s...
Last edited by TSR360; 2 years ago
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RDKGames
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#2
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#2
(Original post by TSR360)
Question (part c): http://prntscr.com/s7jyig
My w/o: http://prntscr.com/s7k0hx

I don't see how I can find the value of t unless I know the value of s...
Vectors \mathbf{a},\mathbf{b} are linearly independent, so it is sufficient for you to compare coefficients of these vectors between both sides and hence solve for s,t simultaneously.
Last edited by RDKGames; 2 years ago
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RDKGames
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(Original post by TSR360)
Question (part c): http://prntscr.com/s7jyig
My w/o: http://prntscr.com/s7k0hx

I don't see how I can find the value of t unless I know the value of s...
A simpler approach, without ever needing to even introduce s, is to observe how the coefficients of \mathbf{a},\mathbf{b} in \overrightarrow{BM} are related (they are just numbers, so the observation should be simple) and impose the same condition on \overrightarrow{BQ} when written in a similar form.

I.e.

\overrightarrow{BM} = -\dfrac{1}{2}\mathbf{b} + \mathbf{a}

note that the coefficient of \mathbf{b} is exactly -1/2 lots of the coefficient of \mathbf{a}.

Impose the same condition on the vector \overrightarrow{BQ} = (t-1)\mathbf{b} + t\mathbf{a}.

Is it clear?
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TSR360
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#4
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(Original post by RDKGames)
A simpler approach, without ever needing to even introduce s, is to observe how the coefficients of \mathbf{a},\mathbf{b} in \overrightarrow{BM} are related (they are just numbers, so the observation should be simple) and impose the same condition on \overrightarrow{BQ} when written in a similar form.

I.e.

\overrightarrow{BM} = -\dfrac{1}{2}\mathbf{b} + \mathbf{a}

note that the coefficient of \mathbf{b} is exactly -1/2 lots of the coefficient of \mathbf{a}.

Impose the same condition on the vector \overrightarrow{BQ} = (t-1)\mathbf{b} + t\mathbf{a}.

Is it clear?
no...
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RDKGames
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#5
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(Original post by TSR360)
no...
Clearly, \overrightarrow{BM} and \overrightarrow{BQ} must be parallel AND go in the same direction, in order to constitute one straight line BQM.

What makes to vectors parallel? You could say one vector being a multiple of the other, but in a more fundamental level, it's when you spot a multiplicative relationship between the \mathbf{i},\mathbf{j} components. You can think of this multiplicative relationship as the ratio between the two components. Any parallel vector must preserve it.

E.g. if \mathbf{a} = -3\mathbf{i} + 6\mathbf{j}, then the \mathbf{j} component is -2 lots of the \mathbf{i} component. This property will not change when you scale the vector \mathbf{a}.

4 \mathbf{a} is parallel to \mathbf{a} but it's -12\mathbf{i} + 24\mathbf{j} and the property discussed above still holds true.

-3 \mathbf{a} is parallel to \mathbf{a} but it's 9 \mathbf{i} - 18\mathbf{j} and the property above still holds true.


Thus, in this question it is much simpler to exploit the property relating the coefficients of \mathbf{b},\mathbf{a} rather than introduce a new variable s.

Clear now?
Last edited by RDKGames; 2 years ago
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