JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 7)
Let the vectors $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$
be such that
$$\left| {\overrightarrow a } \right| = 2$$, $$\left| {\overrightarrow b } \right| = 4$$ and $$\left| {\overrightarrow c } \right| = 4$$. If the projection of
$$\overrightarrow b $$ on $$\overrightarrow a $$ is equal to the projection of $$\overrightarrow c $$ on $$\overrightarrow a $$
and $$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$, then the value of
$$\left| {\overrightarrow a + \vec b - \overrightarrow c } \right|$$ is ___________.
$$\left| {\overrightarrow a } \right| = 2$$, $$\left| {\overrightarrow b } \right| = 4$$ and $$\left| {\overrightarrow c } \right| = 4$$. If the projection of
$$\overrightarrow b $$ on $$\overrightarrow a $$ is equal to the projection of $$\overrightarrow c $$ on $$\overrightarrow a $$
and $$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$, then the value of
$$\left| {\overrightarrow a + \vec b - \overrightarrow c } \right|$$ is ___________.
Answer
6
Explanation
Projection of $$\overrightarrow b $$
on $$\overrightarrow a $$
= Projection of $$\overrightarrow c $$
on $$\overrightarrow a $$
$$ \Rightarrow $$ $${{\overrightarrow b .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = {{\overrightarrow c .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}}$$
$$ \Rightarrow $$ $$\overrightarrow b .\overrightarrow a = \overrightarrow c .\overrightarrow a $$
$$ \because $$ $$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$
$$ \therefore $$ $$\overrightarrow b .\overrightarrow c = 0$$
Let $$\left| {\overrightarrow a + \vec b - \overrightarrow c } \right|$$ = k
Square both sides
k2 = $${{{\left( {\overrightarrow a } \right)}^2}}$$ + $${{{\left( {\overrightarrow b } \right)}^2}}$$ + $${{{\left( {\overrightarrow c } \right)}^2}}$$ + $$2\overrightarrow a .\overrightarrow b $$ - $$2\overrightarrow b .\overrightarrow c $$ - $$2\overrightarrow a .\overrightarrow c $$
$$ \Rightarrow $$ k2 = $${{{\left( {\overrightarrow a } \right)}^2}}$$ + $${{{\left( {\overrightarrow b } \right)}^2}}$$ + $${{{\left( {\overrightarrow c } \right)}^2}}$$
$$ \Rightarrow $$ k2 = 22 + 42 + 42 = 36
$$ \Rightarrow $$ k = 6
$$ \Rightarrow $$ $${{\overrightarrow b .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = {{\overrightarrow c .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}}$$
$$ \Rightarrow $$ $$\overrightarrow b .\overrightarrow a = \overrightarrow c .\overrightarrow a $$
$$ \because $$ $$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$
$$ \therefore $$ $$\overrightarrow b .\overrightarrow c = 0$$
Let $$\left| {\overrightarrow a + \vec b - \overrightarrow c } \right|$$ = k
Square both sides
k2 = $${{{\left( {\overrightarrow a } \right)}^2}}$$ + $${{{\left( {\overrightarrow b } \right)}^2}}$$ + $${{{\left( {\overrightarrow c } \right)}^2}}$$ + $$2\overrightarrow a .\overrightarrow b $$ - $$2\overrightarrow b .\overrightarrow c $$ - $$2\overrightarrow a .\overrightarrow c $$
$$ \Rightarrow $$ k2 = $${{{\left( {\overrightarrow a } \right)}^2}}$$ + $${{{\left( {\overrightarrow b } \right)}^2}}$$ + $${{{\left( {\overrightarrow c } \right)}^2}}$$
$$ \Rightarrow $$ k2 = 22 + 42 + 42 = 36
$$ \Rightarrow $$ k = 6
Comments (0)
