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Question: let \[\overset{\to }{\mathop{a}}\,=\widehat{i}+4\widehat{j}+2\widehat{k},\text{ }\overset{\to }{\mat...

let a→ =i^+4j^+2k^, b→ =3i^−2j^−7k^ and c→ =2i^−j^+4k^.\overset{\to }{\mathop{a}}\,=\widehat{i}+4\widehat{j}+2\widehat{k},\text{ }\overset{\to }{\mathop{b}}\,=3\widehat{i}-2\widehat{j}-7\widehat{k}\text{ and }\overset{\to }{\mathop{c}}\,=2\widehat{i}-\widehat{j}+4\widehat{k}\text{.} find a vectorp→ \overset{\to }{\mathop{p}}\,which is
perpendicular to both a→  and b→ \overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\, and p→ ⋅c→ =18.\overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{c}}\,=18.

Explanation

Solution

In this problem we will find the perpendicular vectorp→ \overset{\to }{\mathop{p}}\,. If two vectors a→  and b→ \overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\, are perpendicular then the dot product of two vectors a→  and b→ \overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\, is zero.
i.e. a→ ⋅b→ =0\overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{b}}\,=0

Complete step by step answer:
Before, start solving the problem let us define a vector. A vector is a quantity which can be completely described using both magnitude and direction.
Mathematically, A vector is a line segment AB with direction from A to B and denoted by AB→ \overset{\to }{\mathop{\text{AB}}}\,
Given that a→ =i^+4j^+2k^, b→ =3i^−2j^−7k^ and c→ =2i^−j^+4k^\overset{\to }{\mathop{a}}\,=\widehat{i}+4\widehat{j}+2\widehat{k},\text{ }\overset{\to }{\mathop{b}}\,=3\widehat{i}-2\widehat{j}-7\widehat{k}\text{ and }\overset{\to }{\mathop{c}}\,=2\widehat{i}-\widehat{j}+4\widehat{k}
To find the vector p→ \overset{\to }{\mathop{p}}\,.
Let p→ =p1i^+p2j^+p3k^\overset{\to }{\mathop{p}}\,={{p}_{1}}\widehat{i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k} be a vector perpendicular to both a→  and b→ \overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{b}}\,and p→ ⋅c→ =18\overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{c}}\,=18
Since p→ =p1i^+p2j^+p3k^\overset{\to }{\mathop{p}}\,=\widehat{{{p}_{1}}i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k} is perpendicular toa→ =i^+4j^+2k^\overset{\to }{\mathop{a}}\,=\widehat{i}+4\widehat{j}+2\widehat{k}
⇒\Rightarrow Dot product of two vectors a→  and p→ \overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{p}}\, is zero
⇒p→ ⋅a→ =0\Rightarrow \overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{a}}\,=0
Now, we will substitute the vectorsa→  and p→ \overset{\to }{\mathop{a}}\,\text{ and }\overset{\to }{\mathop{p}}\,.
⇒(p1i^+p2j^+p3k^)⋅(i^+4j^+2k^)=0\Rightarrow \left( {{p}_{1}}\widehat{i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k} \right)\cdot \left( \widehat{i}+4\widehat{j}+2\widehat{k} \right)=0
By dot product of vectors, we get
⇒p1+4p2+2p3=0....(1)\Rightarrow {{p}_{1}}+4{{p}_{2}}+2{{p}_{3}}=0....(1)
Since p→ =p1i^+p2j^+p3k^\overset{\to }{\mathop{p}}\,=\widehat{{{p}_{1}}i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k} is perpendicular tob→ =3i^−2j^−7k^\overset{\to }{\mathop{b}}\,=3\widehat{i}-2\widehat{j}-7\widehat{k}
⇒\Rightarrow dot product of two vectors b→  and p→ \overset{\to }{\mathop{b}}\,\text{ and }\overset{\to }{\mathop{p}}\, is zero
⇒p→ ⋅b→ =0\Rightarrow \overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{b}}\,=0
Now, we will substitute the vectorsb→  and p→ \overset{\to }{\mathop{b}}\,\text{ and }\overset{\to }{\mathop{p}}\,.
⇒(p1i^+p2j^+p3k^)⋅(3i^−2j^−7k^)=0\Rightarrow \left( {{p}_{1}}\widehat{i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k} \right)\cdot \left( 3\widehat{i}-2\widehat{j}-7\widehat{k} \right)=0
By dot product of vectors, we get
⇒3p1−2p2−7p3=0....(2)\Rightarrow 3{{p}_{1}}-2{{p}_{2}}-7{{p}_{3}}=0....(2)
Also give that p→ ⋅c→ =18\overset{\to }{\mathop{p}}\,\cdot \overset{\to }{\mathop{c}}\,=18
Now, we will substitute the vectorsc→  and p→ \overset{\to }{\mathop{c}}\,\text{ and }\overset{\to }{\mathop{p}}\,.
⇒(p1i^+p2j^+p3k^)⋅(2i^−j^+4k^)=18\Rightarrow \left( {{p}_{1}}\widehat{i}+{{p}_{2}}\widehat{j}+{{p}_{3}}\widehat{k} \right)\cdot \left( 2\widehat{i}-\widehat{j}+4\widehat{k} \right)=18
By dot product of vectors, we get
⇒2p1−p2+4p3=18....(3)\Rightarrow 2{{p}_{1}}-{{p}_{2}}+4{{p}_{3}}=18....(3)
From equation (1) we get
Using the value of p1{{p}_{1}} in equation (2),
⇒3(−4p2−2p3)−2p2−7p3=0\Rightarrow 3\left( -4{{p}_{2}}-2{{p}_{3}} \right)-2{{p}_{2}}-7{{p}_{3}}=0
⇒−12p2−6p3−2p2−7p3=0\Rightarrow -12{{p}_{2}}-6{{p}_{3}}-2{{p}_{2}}-7{{p}_{3}}=0
⇒−14p2−11p3=0\Rightarrow -14{{p}_{2}}-11{{p}_{3}}=0
⇒p2=−1114p3....(5)\Rightarrow {{p}_{2}}=-\dfrac{11}{14}{{p}_{3}}....(5)
Using equation (5) in equation (4) we get,
⇒p1=−4(−1114p3)−2p3\Rightarrow {{p}_{1}}=-4\left( -\dfrac{11}{14}{{p}_{3}} \right)-2{{p}_{3}}
⇒p1=4414p3−2p3\Rightarrow {{p}_{1}}=\dfrac{44}{14}{{p}_{3}}-2{{p}_{3}}
By cross multiplication we get
⇒p1=44p3−28p314\Rightarrow {{p}_{1}}=\dfrac{44{{p}_{3}}-28{{p}_{3}}}{14}
⇒p1=16p314....(6)\Rightarrow {{p}_{1}}=\dfrac{16{{p}_{3}}}{14}....(6)
Using equation (5) and equation (6) in equation (3), we get
⇒2(16p314)−(−1114p3)+4p3=18....(3)\Rightarrow 2\left( \dfrac{16{{p}_{3}}}{14} \right)-\left( -\dfrac{11}{14}{{p}_{3}} \right)+4{{p}_{3}}=18....(3)
⇒32p314+1114p3+4p3=18\Rightarrow \dfrac{32{{p}_{3}}}{14}+\dfrac{11}{14}{{p}_{3}}+4{{p}_{3}}=18
By cross multiplication, we get
⇒32p3+11p3+56p314=18\Rightarrow \dfrac{32{{p}_{3}}+11{{p}_{3}}+56{{p}_{3}}}{14}=18
⇒99p314=18\Rightarrow \dfrac{99{{p}_{3}}}{14}=18
By dividing LHS and RHS by 9, we get
⇒11p314=2\Rightarrow \dfrac{11{{p}_{3}}}{14}=2
By cross multiplication, we get
⇒p3=2811....(7)\Rightarrow {{p}_{3}}=\dfrac{28}{11}....(7)
Using equation (7) in equation (5) and equation (6) we get
⇒p1=1614(2811)\Rightarrow {{p}_{1}}=\dfrac{16}{14}\left( \dfrac{28}{11} \right)
⇒p1=87(2811)\Rightarrow {{p}_{1}}=\dfrac{8}{7}\left( \dfrac{28}{11} \right)
⇒p1=8×411\Rightarrow {{p}_{1}}=\dfrac{8\times 4}{11}
⇒p1=3211\Rightarrow {{p}_{1}}=\dfrac{32}{11}
And p2=−1114(2811){{p}_{2}}=-\dfrac{11}{14}\left( \dfrac{28}{11} \right)
p2=−2{{p}_{2}}=- 2
p2=−2{{p}_{2}}=-2
Hencep1=3211{{p}_{1}}=\dfrac{32}{11}, p2=−2{{p}_{2}}=-2 andp3=2811{{p}_{3}}=\dfrac{28}{11}.
Therefore perpendicular vector is p→ =3211i^−2j^+2811k^\overset{\to }{\mathop{p}}\,=\dfrac{32}{11}\widehat{i}- 2\widehat{j}+\dfrac{28}{11}\widehat{k}

Note:
In this problem, one knows that if two vectors are perpendicular then their dot product is zero. Alternative, to find the coordinate of vector p→ \overset{\to }{\mathop{p}}\, i.e. p1, p2 and p3{{\text{p}}_{1}}\text{, }{{\text{p}}_{\text{2}}}\text{ and }{{\text{p}}_{3}} we can solve equation (1), equation (2) and equation (3) using matrices. By converting the system of equations in matrix form.