
How to find a vector A that has the same directions as \[\left\langle -8,7,8 \right\rangle \]but has length 3?
Answer
573.6k+ views
Hint: Here we have to find a vector A that has the same directions as \[\left\langle -8,7,8 \right\rangle \]but has length 3. This problem is based on scaling and similarity. we have to know that any vector that has the same direction as \[\left\langle -8,7,8 \right\rangle \]has all coordinates proportional to the given vector. Now we can describe directions by coordinates, from the coordinates, we have to find a scaling factor that leads to a vector with length 3.
Complete step by step answer:
We know that the given direction is \[\left\langle -8,7,8 \right\rangle \].
Now we can write the direction with coordinates, we get \[\left\langle -8f,7f,8f \right\rangle \], where f is the scaling factor.
We also know that the length of the vector with coordinates \[\left\langle af,bf,cf \right\rangle \]is equal to,
\[\sqrt{{{a}^{2}}{{f}^{2}}+{{b}^{2}}{{f}^{2}}+{{c}^{2}}{{f}^{2}}}=f\times \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]
we have the direction, whose length of the vector with coordinates \[\left\langle -8f,7f,8f \right\rangle \]is equal to,
\[\begin{align}
& \Rightarrow \sqrt{{{\left( -8 \right)}^{2}}{{f}^{2}}+{{\left( 7 \right)}^{2}}{{f}^{2}}+{{\left( 8 \right)}^{2}}{{f}^{2}}} \\
& \Rightarrow f\times \sqrt{64+49+64} \\
& \Rightarrow f\sqrt{177} \\
\end{align}\]
Since, we have length is equal to 3, we can write
\[\begin{align}
& \Rightarrow f\sqrt{177}=3 \\
& \Rightarrow f=\dfrac{3}{\sqrt{177}} \\
\end{align}\]
Here, we have found the scaling vector and we can multiply with the direction \[\left\langle -8,7,8 \right\rangle \]to get the coordinates of the vector A,
Therefore, the vector A is \[\left( \dfrac{-24}{\sqrt{177}},\dfrac{21}{\sqrt{177}},\dfrac{24}{\sqrt{177}} \right)\]
Note: Students make mistakes in finding the length of the coordinates, it is easier for students to remember the formula to find the length of the coordinates. In this problem, the step where we multiplied the given length 3, is the place where students may make mistakes. This problem requires the understanding of scaling and similarity, to solve these types of problems.
Complete step by step answer:
We know that the given direction is \[\left\langle -8,7,8 \right\rangle \].
Now we can write the direction with coordinates, we get \[\left\langle -8f,7f,8f \right\rangle \], where f is the scaling factor.
We also know that the length of the vector with coordinates \[\left\langle af,bf,cf \right\rangle \]is equal to,
\[\sqrt{{{a}^{2}}{{f}^{2}}+{{b}^{2}}{{f}^{2}}+{{c}^{2}}{{f}^{2}}}=f\times \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]
we have the direction, whose length of the vector with coordinates \[\left\langle -8f,7f,8f \right\rangle \]is equal to,
\[\begin{align}
& \Rightarrow \sqrt{{{\left( -8 \right)}^{2}}{{f}^{2}}+{{\left( 7 \right)}^{2}}{{f}^{2}}+{{\left( 8 \right)}^{2}}{{f}^{2}}} \\
& \Rightarrow f\times \sqrt{64+49+64} \\
& \Rightarrow f\sqrt{177} \\
\end{align}\]
Since, we have length is equal to 3, we can write
\[\begin{align}
& \Rightarrow f\sqrt{177}=3 \\
& \Rightarrow f=\dfrac{3}{\sqrt{177}} \\
\end{align}\]
Here, we have found the scaling vector and we can multiply with the direction \[\left\langle -8,7,8 \right\rangle \]to get the coordinates of the vector A,
Therefore, the vector A is \[\left( \dfrac{-24}{\sqrt{177}},\dfrac{21}{\sqrt{177}},\dfrac{24}{\sqrt{177}} \right)\]
Note: Students make mistakes in finding the length of the coordinates, it is easier for students to remember the formula to find the length of the coordinates. In this problem, the step where we multiplied the given length 3, is the place where students may make mistakes. This problem requires the understanding of scaling and similarity, to solve these types of problems.
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