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Question: If \(\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat {4i} - \w...

If a=i^+j^+k^,b=4i^2j^+3k^,c=i^2j^+k^\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat {4i} - \widehat {2j} + \widehat {3k},\overrightarrow c = \widehat i - \widehat {2j} + \widehat k . Find a vector of magnitude 6 units which is parallel to the vector 2ab3c2\overrightarrow a - \overrightarrow b - 3\overrightarrow c .

Explanation

Solution

We are given three vectors and let r=2ab3c\overrightarrow r = 2\overrightarrow a - \overrightarrow b - 3\overrightarrow c and substituting the given vectors we get the new vector r and in order to find the unit vector we use the formula r^=rr\widehat r = \dfrac{{\overrightarrow r }}{{\left| {\overrightarrow r } \right|}}and the magnitude formula is given as (coefficient of i)2+(coefficient of j)2+(coefficient of k)2\sqrt {{{\left( {{\text{coefficient of i}}} \right)}^2} + {{\left( {{\text{coefficient of j}}} \right)}^2} + {{\left( {{\text{coefficient of k}}} \right)}^2}} and after obtaining the unit vector the required vector is obtained by multiplying 6 with the unit vector.

Complete step by step solution:
Let r=2ab3c\overrightarrow r = 2\overrightarrow a - \overrightarrow b - 3\overrightarrow c
We are given that a=i^+j^+k^,b=4i^2j^+3k^,c=i^2j^+k^\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat {4i} - \widehat {2j} + \widehat {3k},\overrightarrow c = \widehat i - \widehat {2j} + \widehat k
Substituting this in we get
r=2(i^+j^+k^)(4i^2j^+3k^)3(i^2j^+k^) r=2i^+2j^+2k^4i^+2j^3k^3i^+6j^3k^ r=5i^+10j^4k^  \Rightarrow \overrightarrow r = 2\left( {\widehat i + \widehat j + \widehat k} \right) - \left( {\widehat {4i} - \widehat {2j} + \widehat {3k}} \right) - 3\left( {\widehat i - \widehat {2j} + \widehat k} \right) \\\ \Rightarrow \overrightarrow r = 2\widehat i + \widehat {2j} + 2\widehat k - \widehat {4i} + \widehat {2j} - \widehat {3k} - 3\widehat i + 6\widehat j - 3\widehat k \\\ \Rightarrow \overrightarrow r = - 5\widehat i + 10\widehat j - \widehat {4k} \\\
Now the unit vector is given by the formula
r^=rr\Rightarrow \widehat r = \dfrac{{\overrightarrow r }}{{\left| {\overrightarrow r } \right|}} ……….(1)
Where r\overrightarrow r is the given vector and r\left| {\overrightarrow r } \right|is the magnitude of r\overrightarrow r
The magnitude of a vector is given by (coefficient of i)2+(coefficient of j)2+(coefficient of k)2\sqrt {{{\left( {{\text{coefficient of i}}} \right)}^2} + {{\left( {{\text{coefficient of j}}} \right)}^2} + {{\left( {{\text{coefficient of k}}} \right)}^2}}
Therefore the modulus of is given as

r=(5)2+(10)2+(4)2 r=25+100+16=141 r=141  \Rightarrow \left| {\overrightarrow r } \right| = \sqrt {{{\left( { - 5} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( { - 4} \right)}^2}} \\\ \Rightarrow \left| {\overrightarrow r } \right| = \sqrt {25 + 100 + 16} = \sqrt {141} \\\ \Rightarrow \left| {\overrightarrow r } \right| = \sqrt {141} \\\

Using this in (1) we get
r^=5i^+10j^4k^141\Rightarrow \widehat r = \dfrac{{ - 5\widehat i + 10\widehat j - \widehat {4k}}}{{\sqrt {141} }}
Since we are given the magnitude of the required vector is 6
The vector parallel to r with magnitude 6 is given by 6×r^6\times \widehat r
6×5i^+10j^4k^141 30i^+60j^24k^141  \Rightarrow 6\times \dfrac{{ - 5\widehat i + 10\widehat j - \widehat {4k}}}{{\sqrt {141} }} \\\ \Rightarrow \dfrac{{ - 30\widehat i + 60\widehat j - 24\widehat k}}{{\sqrt {141} }} \\\

Therefore the required vector is 30i^+60j^24k^141\dfrac{{ - 30\widehat i + 60\widehat j - 24\widehat k}}{{\sqrt {141} }}.

Note :
Vectors are parallel if they have the same direction. Both components of one vector must be in the same ratio to the corresponding components of the parallel vector.
A unit vector is a vector of length 1, sometimes also called a direction vector.