
If the coordinates of the point A, B, C be $( - 1,3,2)$ , $(2,3,5)$ and $(3,5, - 2)$ respectively, then what is the value of angle A?
Answer
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Hint: Determine the direction ratios of the lines $AB$ and $AC$. Now, use the property of the dot product to evaluate the value of the angle between the two lines, that is, angle A. Angles between two straight lines is the angle between their direction vectors. Thus, we have $\cos A = \dfrac{{{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}}}{{\sqrt {{l_1}^2 + {m_1}^2 + {n_1}^2} \sqrt {{l_2}^2 + {m_2}^2 + {n_2}^2} }}$ , where $({l_1},{m_1},{n_1})$ and $({l_2},{m_2},{n_2})$ are the direction ratios between the two lines respectively.
Formula Used: $\cos A = \dfrac{{{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}}}{{\sqrt {{l_1}^2 + {m_1}^2 + {n_1}^2} \sqrt {{l_2}^2 + {m_2}^2 + {n_2}^2} }}$ , where $({l_1},{m_1},{n_1})$ and $({l_2},{m_2},{n_2})$ are the direction ratios between the two lines respectively.
Complete step by step Solution:
Given coordinates are:
$A( - 1,3,2)$ , $B(2,3,5)$ and $C(3,5, - 2)$
Evaluating the direction ratios of the line $AB$,
${l_1} = 2 - ( - 1) = 3$
${m_1} = 3 - 3 = 0$
${n_1} = 5 - 2 = 3$
Thus, the direction ratios of line $AB$ are $({l_1},{m_1},{n_1}) = (3,0,3)$ … (1)
Similarly, evaluating the direction ratios of the line AC,
${l_2} = 3 - ( - 1) = 4$
${m_2} = 5 - 3 = 2$
${n_2} = - 2 - 2 = - 4$
Thus, the direction ratios of the lien $AC$ are $\left( {{l_2},{m_2},{n_2}} \right) = (4,2, - 4)$ … (2)
Now, the angle between these two lines will be calculated using their direction ratios:
$\cos A = \dfrac{{{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}}}{{\sqrt {{l_1}^2 + {m_1}^2 + {n_1}^2} \sqrt {{l_2}^2 + {m_2}^2 + {n_2}^2} }}$
Substituting the values of the direction ratios from (1) and (2)
$\cos A = \dfrac{{(3)(4) + (0)(2) + (3)( - 4)}}{{\sqrt {{3^2} + {0^2} + {3^2}} \sqrt {{4^2} + {2^2} + {{( - 4)}^2}} }}$
On simplifying further,
$\cos A = 0$
As an angle of a triangle cannot be greater than or equal to ${180^ \circ }$, therefore, taking the cosine inverse, we have:
$A = {90^ \circ }$
Hence, the value of angle A is ${90^ \circ }$.
Note: Using dot product is easier to evaluate the angles between two lines or two vectors. Dot product gives us the angle between them in the form of its cosine and it also involves much easier calculations than the cross product.
Formula Used: $\cos A = \dfrac{{{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}}}{{\sqrt {{l_1}^2 + {m_1}^2 + {n_1}^2} \sqrt {{l_2}^2 + {m_2}^2 + {n_2}^2} }}$ , where $({l_1},{m_1},{n_1})$ and $({l_2},{m_2},{n_2})$ are the direction ratios between the two lines respectively.
Complete step by step Solution:
Given coordinates are:
$A( - 1,3,2)$ , $B(2,3,5)$ and $C(3,5, - 2)$
Evaluating the direction ratios of the line $AB$,
${l_1} = 2 - ( - 1) = 3$
${m_1} = 3 - 3 = 0$
${n_1} = 5 - 2 = 3$
Thus, the direction ratios of line $AB$ are $({l_1},{m_1},{n_1}) = (3,0,3)$ … (1)
Similarly, evaluating the direction ratios of the line AC,
${l_2} = 3 - ( - 1) = 4$
${m_2} = 5 - 3 = 2$
${n_2} = - 2 - 2 = - 4$
Thus, the direction ratios of the lien $AC$ are $\left( {{l_2},{m_2},{n_2}} \right) = (4,2, - 4)$ … (2)
Now, the angle between these two lines will be calculated using their direction ratios:
$\cos A = \dfrac{{{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}}}{{\sqrt {{l_1}^2 + {m_1}^2 + {n_1}^2} \sqrt {{l_2}^2 + {m_2}^2 + {n_2}^2} }}$
Substituting the values of the direction ratios from (1) and (2)
$\cos A = \dfrac{{(3)(4) + (0)(2) + (3)( - 4)}}{{\sqrt {{3^2} + {0^2} + {3^2}} \sqrt {{4^2} + {2^2} + {{( - 4)}^2}} }}$
On simplifying further,
$\cos A = 0$
As an angle of a triangle cannot be greater than or equal to ${180^ \circ }$, therefore, taking the cosine inverse, we have:
$A = {90^ \circ }$
Hence, the value of angle A is ${90^ \circ }$.
Note: Using dot product is easier to evaluate the angles between two lines or two vectors. Dot product gives us the angle between them in the form of its cosine and it also involves much easier calculations than the cross product.
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