Question

Let a three-dimensional vector $\overrightarrow{V}$ satisfy the condition, $2\overrightarrow{V}+\overrightarrow{V}\times (i+2j)=2i+k$. If $3\left| \overrightarrow{V} \right|=\sqrt{m}$ .​ Then find the value of m.

Answer 1

Hint: To solve this question we need to know about the products of vectors, which are dot products and cross products. We will apply the formula for solving the expression. They are $\overrightarrow{a}\bullet \overrightarrow{b}=\left| a \right|\left| b \right|\cos \theta $ and $\overrightarrow{a}\times \overrightarrow{b}=\left| a \right|\left| b \right|\sin \theta $ . We will also use Scalar Triple Product to solve the problem with less calculation.

Complete step by step answer:
The question ask us to find the value of “m” if one of the relation giver is $3\left| \overrightarrow{V} \right|=\sqrt{m}$ while the other expression is $2\overrightarrow{V}+\overrightarrow{V}\times (i+2j)=2i+k$. In the first step to solve the expression we will choose for the process in which will give the least possible calculation required. The expression give to us is:
$\Rightarrow 2\overrightarrow{V}+\overrightarrow{V}\times (i+2j)=2i+k$…….(i)
Since, we do not know about the vector $\overrightarrow{V}$, so solving for cross product becomes difficult. This is the reason we will take the help of Scalar Triple Product which states that when a same vector is cross multiplied to another vector and then dot product is found with the vector itself then the expression results to zero. Mathematically written as $\left( a\times b \right)\bullet a=0$. In a similar manner we will multiply each term in the above given expression with $(i+2j)$and find the dot product. On doing this we get:
\[\Rightarrow 2\overrightarrow{V}\bullet (i+2j)+\left( \overrightarrow{V}\times (i+2j) \right)\bullet (i+2j)=\left( 2i+k \right)\bullet (i+2j)\]
\[\Rightarrow 2\overrightarrow{V}\bullet (i+2j)+0=2+0+0\]
To the term in Left Hand Side we will apply $\overrightarrow{a}\bullet \overrightarrow{b}=\left| a \right|\left| b \right|\cos \theta $. On using formula we get:
\[\Rightarrow \left| \overrightarrow{V} \right|\left| i+2j \right|\cos \theta =1\]
We will square the whole expression so that the value is not in root. On doing this we get:
\[\Rightarrow {{\left| \overrightarrow{V} \right|}^{2}}{{\left| i+2j \right|}^{2}}{{\cos }^{2}}\theta ={{1}^{2}}\]
\[\Rightarrow {{\left| \overrightarrow{V} \right|}^{2}}{{\left( \sqrt{{{1}^{2}}+{{2}^{2}}} \right)}^{2}}{{\cos }^{2}}\theta =1\]
\[\Rightarrow {{\left| \overrightarrow{V} \right|}^{2}}5{{\cos }^{2}}\theta ={{1}^{2}}\]
Now, to solve this question we will apply the trigonometric identity which states ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. On applying we get:
\[\Rightarrow {{\left| \overrightarrow{V} \right|}^{2}}5\left( 1-{{\sin }^{2}}\theta \right)=1\]
\[\Rightarrow 5{{\left| \overrightarrow{V} \right|}^{2}}-5{{\left| \overrightarrow{V} \right|}^{2}}{{\sin }^{2}}\theta =1\]
\[\Rightarrow 5{{\left| \overrightarrow{V} \right|}^{2}}{{\sin }^{2}}\theta =1-5{{\left| \overrightarrow{V} \right|}^{2}}\]…….. (ii)
The above equation is not sufficient for us to find the value of “m”. So we will square the equation (i), to get another relation. On squaring we get:
 $\Rightarrow {{\left\{ 2\overrightarrow{V}+\overrightarrow{V}\times (i+2j) \right\}}^{2}}={{\left( 2i+k \right)}^{2}}$
\[\Rightarrow 4{{\left| \overrightarrow{V} \right|}^{2}}+{{\left| \overrightarrow{V} \right|}^{2}}{{(i+2j)}^{2}}{{\sin }^{2}}\theta +2\overrightarrow{V}\bullet \left( \overrightarrow{V}\times (i+2j) \right)=4+1\]
\[\Rightarrow 4{{\left| \overrightarrow{V} \right|}^{2}}+{{\left| \overrightarrow{V} \right|}^{2}}5{{\sin }^{2}}\theta +0=5\]
\[\Rightarrow 5{{\left| \overrightarrow{V} \right|}^{2}}{{\sin }^{2}}\theta =5-4{{\left| \overrightarrow{V} \right|}^{2}}\]
We will substitute \[5{{\left| \overrightarrow{V} \right|}^{2}}{{\sin }^{2}}\theta \] with the equation (ii). On doing this we get:
\[\Rightarrow 4{{\left| \overrightarrow{V} \right|}^{2}}+5{{\left| \overrightarrow{V} \right|}^{2}}-1=5\]
\[\Rightarrow 9{{\left| \overrightarrow{V} \right|}^{2}}=5+1=6\]
On finding the square root of the term we get:
\[\Rightarrow 3\left| \overrightarrow{V} \right|=\sqrt{6}\]
Comparing the expression $3\left| \overrightarrow{V} \right|=\sqrt{m}$ given in the question we see that $m=6$.
$\therefore $ The value of “m” in $3\left| \overrightarrow{V} \right|=\sqrt{m}$ is $6$.

Note: To solve the problem on vectors, we should keep in mind that the dot product of two similar vectors is zero, while the cross product of two perpendicular vectors is zero. Both the above conditions could be found with the help of the formula for dot and cross product respectively.