Find the value of x when two vectors are parallel or equal in magnitude

Key concept: This is a classic vector-components question. One part uses proportional components for parallel vectors, and the other uses the formula for vector length.

Step by step

Let

$\mathbf a=7\mathbf i+6\mathbf j=(7,6),\qquad \mathbf b=2\mathbf i+x\mathbf j=(2,x).$

a) When is $\mathbf a$ parallel to $\mathbf b$?

For parallel vectors, one is a scalar multiple of the other:

$(7,6)=k(2,x).$

Compare the $i$-components:

$7=2k \implies k=\frac72.$

Now compare the $j$-components:

$6=kx=\frac72x.$

So

$x=\frac{12}{7}.$

b) When do $\mathbf a$ and $\mathbf b$ have the same magnitude?

First find $|\mathbf a|$:

$|\mathbf a|=\sqrt{7^2+6^2}=\sqrt{85}.$

Also

$|\mathbf b|=\sqrt{2^2+x^2}=\sqrt{4+x^2}.$

Set them equal:

$\sqrt{4+x^2}=\sqrt{85}$

so

$4+x^2=85,$

$x^2=81,$

$x=\pm 9.$

Therefore:

• for parallel vectors, $\boxed{x=\frac{12}{7}}$
• for equal magnitude, $\boxed{x=9\text{ or }x=-9}$

Pitfall alert

For parallel vectors, don’t compare just one component and stop there; both components must match the same scalar factor. For magnitude, watch the square root step carefully and remember both positive and negative values of $x$ can work.

Try different conditions

If the question asked for $\mathbf b$ to be perpendicular to $\mathbf a$, you would use the dot product: $7\cdot 2+6x=0$. That would give a different value of $x$. If the magnitude condition changed to $|\mathbf b|=10$, then you would solve $4+x^2=100$ instead.

Further reading

parallel vectors, magnitude of a vector, component form