Question

How do you find a vector of length 10 that is oppositely directed to $u=3i-4j$?

Answer 1

Hint: We first find the formula for the opposite vector of the given vector $u=3i-4j$. Then using the modulus value for the opposite vector we multiply the vector with 2 to find the vector of length 10 that is oppositely directed to $\overrightarrow{u}=3i-4j$.

Complete step by step solution:
Let us assume the vector of length 10 that is oppositely directed to $\overrightarrow{u}=3i-4j$ is $\overrightarrow{v}$. So, $\left| \overrightarrow{v} \right|=10$.
We know that any vector directly opposite to a vector $\overrightarrow{a}$ will be $-\left( \overrightarrow{a} \right)$.
This means that the individual signs of the coefficients change.
Following the same process, we get that the vector that is oppositely directed to $\overrightarrow{u}=3i-4j$ is
$\overrightarrow{x}=-\left( 3i-4j \right)=-3i+4j$.
But it is also mentioned that the length of the vector is 10 units.
The length of any vector $\overrightarrow{a}=mi+nj$ is $\left| \overrightarrow{a} \right|=\sqrt{{{m}^{2}}+{{n}^{2}}}$.
For our oppositely directed vector $\overrightarrow{x}=-3i+4j$, the length is $\left| \overrightarrow{x} \right|=\sqrt{{{\left( -3 \right)}^{2}}+{{4}^{2}}}=\sqrt{9+16}=\sqrt{25}=5$ units.
We need a vector of length 10 units. The ratio in which we have to multiply the vector $\overrightarrow{x}=-3i+4j$ to get a vector of length 10 is $\dfrac{10}{5}=2$.
Therefore, $\overrightarrow{v}=2\overrightarrow{x}$. Multiplying we 2 we get $\overrightarrow{v}=2\left( -3i+4j \right)=-6i+8j$.

The required vector is $-6i+8j$.

Note: We need to remember that the modulus value of a vector is only dependent on the coefficients of the vector. That’s why we could use the relation of $\overrightarrow{v}=2\overrightarrow{x}$ for the length value of 10.