In the product
$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})=\mathrm{q}\overrightarrow{\mathrm{v}}\times \left(\mathrm{B}\widehat{\mathrm{i}}+\mathrm{B}\widehat{\mathrm{j}}+{\mathrm{B}}_0\hat{\mathrm{k}}\right)$
For q = 1 and $\overrightarrow{\mathrm{v}}=2\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{F}}=4\widehat{\mathrm{i}}-20\widehat{\mathrm{j}}+12\hat{\mathrm{k}}$
What will be the complete expression for $\overrightarrow{\mathrm{B}}$ ?

$\begin{array}{l}\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}})\\ \overrightarrow{\mathrm{F}}\cdot \overrightarrow{\mathrm{B}}=0\\ \overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}}=\widehat{\mathrm{i}}\left[4{\mathrm{B}}_0-6\mathrm{B}\right]+\widehat{\mathrm{j}}\left[6\mathrm{B}-2{\mathrm{B}}_0\right]+\widehat{\mathrm{k}}[2\mathrm{B}-4\mathrm{B}]\\ =\widehat{\mathrm{i}}\left[4{\mathrm{B}}_0-6\mathrm{B}\right]+\widehat{\mathrm{j}}\left(6\mathrm{B}-2{\mathrm{B}}_0\right)-\widehat{\mathrm{k}}\left(2\mathrm{B}\right)\end{array}$
4B₀ - 6B = 4 ...(1)
6B - 2B₀ = -20 ...(2)
-2B = 12 ...(3)
Solving (1), (2) and (3) we get
B = -6 and B₀ = -8