Question

Given $A = ai + bj + ck,B = di + 3j + 4k$ and \[C = 3i + j - 2k\]. If the vectors \[A,B\] and $C$ form a triangle such that \[A = B + C\] and area $(\Delta ABC) = 5\sqrt {6,} $ then

A. $a = - 8,b = - 4,c = - 2,d = - 11$
B. $a = - 8,b = 4,c = - 2,d = - 11$
C. $a = - 8,b = 4,c = 2,d = - 11$
D. None of the above

Answer 1

Hint- In order to solve this question first we will use the vector sum of two vectors and we will get the value of a, b and c by comparing two vectors further we will evaluate the area of triangle by using the concept as area of triangle formed is equal to \[1/2\] times the magnitude of cross product of the \[2\] vectors forming the triangle with the vectors being continuous.

Complete step-by-step answer:
Given that
$
  A = ai + bj + ck \\
  B = di + 3j + 4k \\
$
And \[C = 3i + j - 2k\]
Here \[A,B,C\] are the vectors which represent the sides of the triangle \[ABC.\]
It is given that \[A = B + C\]
Now we will substitute the value of \[A,B\] and $C$in above equation so we have
\[ai + bj + ck = (d + 3)i + 4j + 2k\]
By comparing the components of \[i\] vector, \[j\] vector and \[k\] vector , we have
\[a = d + 3,b = 4,c = 2\]
As we know that the area of the triangle formed is equal to \[1/2\] times the magnitude of the cross product of the \[2\] vectors forming the triangle with the vectors being continuous.
So first we will evaluate the cross product of any two given vectors as \[B{\text{ }} \times {\text{ }}C\]
\[ \Rightarrow \] \[\left| {\begin{array}{*{20}{c}}
  i&j&k \\
  d&3&4 \\
  3&1&{ - 2}
\end{array}} \right| = - 10i + (2d + 12)j + (d - 9)k\]
As we know that magnitude of a vector \[x{\text{ }} = {\text{ }}pi{\text{ }} + {\text{ }}qj{\text{ }} + {\text{ }}rk\] is given as

\[\left| x \right| = \sqrt {{p^2} + {q^2} + {r^2}} \]

By applying above formula we have

\[\sqrt {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}} \]\[\left| {B \times C} \right| = \sqrt {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}} \]


Therefore the magnitude of cross vector of \[B{\text{ }} \times {\text{ }}C\] is \[\sqrt {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}} \]
Now we will evaluate the value of area of triangle , so from above property we will write the formula as
Area of \[\Delta ABC{\text{ }} = \]\[\dfrac{1}{2}\left| {B \times C} \right|\]
Substitute the value of magnitude of vector \[B{\text{ }} \times {\text{ }}C\] in above formula we have
\[\dfrac{1}{2}\sqrt {\left[ {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}} \right]} \]
Area of triangle \[ABC.\] is given as
\[\Delta ABC{\text{ }} = \] $5\sqrt 6 $
So equation becomes as
\[\dfrac{1}{2}\sqrt {\left[ {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}} \right]} = 5\sqrt 6 \]
\[ \Rightarrow \sqrt {5{d^2} + 30d + 325} = 10\sqrt 6 \]
Further by taking square of both the sides we have
\[5{d^2} + 30d + 325 = 600\]
Now by solving quadratic equation we get two solution of \[d\] as
\[
  5{d^2} + 30d - 275 = 0 \\
   \Rightarrow {d^2} + 6d - 55 = 0 \\
\]
\[
   \Rightarrow (d + 11)(d - 5) = 0 \\
   \Rightarrow d = 5, - 11 \\
\]
By considering value of \[d{\text{ }} = {\text{ }}5\]
Because
\[a = d + 3\]
Put the value of \[d\] in above equation
So
\[
  a{\text{ }} = {\text{ }}5 + 3 \\
  a = 8 \\
\]
Now by taking value of \[d\] as \[ - 11\]
\[a = d + 3\]
Put the value of \[d\] in above equation
So
\[
  a = - 11 + 3 \\
  a = - 8 \\
\]
Therefore if \[d{\text{ }} = {\text{ }}5\] then \[a{\text{ }} = {\text{ }}8{\text{ }},{\text{ }}b{\text{ }} = {\text{ }}4\;\]and \[c{\text{ }} = {\text{ }}2.\]
Or if \[d{\text{ }} = {\text{ }} - 11\] then \[a{\text{ }} = {\text{ }} - 8{\text{ }},{\text{ }}b{\text{ }} = {\text{ }}4\] and \[c{\text{ }} = {\text{ }}2.\]
Hence the correct answer is option C.

Note- In algebra, a quadratic equation is any equation that can be rearranged in standard form as \[a{x^2} + bx + c = 0\] where \[x\] represents an unknown, and \[a,{\text{ }}b,{\text{ }}and{\text{ }}c\] represent known numbers, where \[a \ne 0\]. If\[a = 0\], then the equation is linear, not quadratic, as there is no term.