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Question: Show that the points \(A,B,C\) with position vectors \(2\widehat i - \widehat j + \widehat k,\wideha...

Show that the points A,B,CA,B,C with position vectors 2i^−j^+k^,i^−3j^−5k^2\widehat i - \widehat j + \widehat k,\widehat i - 3\widehat j - 5\widehat k and 3i^−4j^−4k^3\widehat i - 4\widehat j - 4\widehat k respectively, are the vertices of a right angled triangle. Hence find the area of the triangle.

Explanation

Solution

In this question the position vectors of the vertices are given. So, first try to find out the length of all sides of the triangle. Using Pythagoras theorem if it follows Pythagoras theorem then it will be proved that the triangle is a right angled triangle. If it is a right angled triangle then find out the area of the triangle.
Given: The points A,B,CA,B,C with position vectors
A→=2i^−j^+k^\overrightarrow A = 2\widehat i - \widehat j + \widehat k
B→=i^−3j^−5k^\overrightarrow B = \widehat i - 3\widehat j - 5\widehat k
C→=3i^−4j^−4k^\overrightarrow C = 3\widehat i - 4\widehat j - 4\widehat k

Step-by-step solution: Since, the position vectors of all the vertex of triangles are given so first we try to find out AB‾,BC‾\overline {AB} ,\overline {BC} and of the given vector.
We know that AB‾=B‾−A‾\overline {AB} = \overline B - \overline A
Similarly, BC‾=C‾−B‾\overline {BC} = \overline C - \overline B
CA‾=A‾−C‾\overline {CA} = \overline A - \overline C
∴AB‾=(1−2)i^+(1−3)j^+(−1−5)k^ =−i^−2j^−6k^ BC‾=C‾−B‾=(3−1)i^+(−4+3)j^+(−4+5)k^ =2i^−j^+k^ CA‾=A‾−C‾=−i^+3j^+5k^  \therefore \overline {AB} = (1 - 2)\widehat i + (1 - 3)\widehat j + ( - 1 - 5)\widehat k \\\ = - \widehat i - 2\widehat j - 6\widehat k \\\ \overline {BC} = \overline C - \overline B = (3 - 1)\widehat i + ( - 4 + 3)\widehat j + ( - 4 + 5)\widehat k \\\ = 2\widehat i - \widehat j + \widehat k \\\ \overline {CA} = \overline A - \overline C = - \widehat i + 3\widehat j + 5\widehat k \\\
∴\therefore Length of ∣AB‾∣=(−1)2+(−2)2+(−6)2=1+4+36=41\left| {\overline {AB} } \right| = \sqrt {{{( - 1)}^2} + {{( - 2)}^2} + {{( - 6)}^2}} = \sqrt {1 + 4 + 36} = \sqrt {41} (Formula used: ai^+bj^+ck^a\widehat i + b\widehat j + c\widehat k is given then magnitude will be a2+b2+c2\sqrt {{a^2} + {b^2} + {c^2}} )
∴\therefore Length of ∣BC‾∣=(2)2+(−1)2+(1)2=4+1+1=6\left| {\overline {BC} } \right| = \sqrt {{{(2)}^2} + {{( - 1)}^2} + {{(1)}^2}} = \sqrt {4 + 1 + 1} = \sqrt 6
∴\therefore Length of ∣AC‾∣=(−1)2+(3)2+(5)2=1+9+25=35\left| {\overline {AC} } \right| = \sqrt {{{( - 1)}^2} + {{(3)}^2} + {{(5)}^2}} = \sqrt {1 + 9 + 25} = \sqrt {35}
From the above obtained length it show that
∴(AB)2=(BC)2+(AC)2\therefore {(AB)^2} = {(BC)^2} + {(AC)^2}
⇒(41)2=(6)2+(35)2 ⇒41=6+35 ∴41=41  \Rightarrow {(\sqrt {41} )^2} = {(6)^2} + {(35)^2} \\\ \Rightarrow 41 = 6 + 35 \\\ \therefore 41 = 41 \\\
Hence, the given triangle ABCABC is a right angled triangle.
∵\because Base = ∣BC‾∣\left| {\overline {BC} } \right| Height = ∣CA‾∣\left| {\overline {CA} } \right|
Now, area of triangle ABC=12×Base×HeightABC = \dfrac{1}{2} \times Base \times Height

Finding cross product
BC‾×CA‾\overline {BC} \times \overline {CA}

\overline {BC} \times \overline {CA} = \left( {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ 2&{ - 1}&1 \\\ { - 1}&3&5 \end{array}} \right) \\\ = \widehat i( - 5 - 3) - \widehat j(10 + 1) + \widehat k(6 - 1) \\\ = - 8\widehat i - 11\widehat j + 5\widehat k \\\

∣BC‾−CA‾∣=64+121+25=210\left| {\overline {BC} - \overline {CA} } \right| = \sqrt {64 + 121 + 25} = \sqrt {210}
∴\therefore Area of triangle ABC=12210ABC = \dfrac{1}{2}\sqrt {210} unit square
=2102= \dfrac{{\sqrt {210} }}{2}

Note: In this question vector order arrangement must be the same otherwise it will mislead the result. When we try to find out the cross product for the area then the sign should be according to proper instruction.