Question

(a) To determine

To find: A nonzero vector orthogonal to the plane through the points \({\bf{P}}\), \({\bf{Q}}\) and \(R\).

(b) To determine

To find: The area of triangle \({\bf{PQ}}R\).

Short answer

(a)The nonzero vector that runs through the points \(P\), \(Q\) and \(R\) is \(\langle 13, - 14,5\rangle \) orthogonal to the plane.

(b)The area of triangle \(P{\rm{ }}Q{\rm{ }}R\) is \(\frac{1}{2}(\sqrt {390} )\).

Step-by-step solution

Concept of Nonzero vector, Orthogonal vectors, cross product and Area of Triangle with the help of Formula.

A vector with at least one non-zero entry, at least in Rn or Cn, is called a non-zero vector. In general, a non-zero vector is one that is not the vector space's identity element for addition.

Two vectors are said to be orthogonal if they are perpendicular to one another. The dot product of the two vectors, in other words, is zero.

Formula

(i) Write the expression for cross product between\(a\)and\(b\)vectors.

\(a \times b = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|(1)\)

(ii) Consider the expression for area of the triangle in vector coordinate.

Area of triangle \( = \frac{1}{2}|a \times b|\left( 2 \right)\)

calculating non zero orthogonal vector points with cross product formula.

(a)

\(P(0, - 2,0),Q(4,1, - 2)andR(5,3,1)\)

The vectors \(\overrightarrow {PQ} \times \overrightarrow {PR} \) is perpendicular to both \(\overrightarrow {PQ} \) and \(\overrightarrow {PR} \), therefore perpendicular to the plane through \(P(0, - 2,0),Q(4,1, - 2)\) and \(R(5,3,1)\).

Find \(\overrightarrow {PQ} \).

\(\begin{array}{l}\overrightarrow {PQ} = Q(4,1,2) - P(0, - 2,0)\\\overrightarrow {PQ} = \langle (4 - 0),(1 + 2),( - 2 - 0)\rangle \\\overrightarrow {PQ} = \langle 4,3, - 2\rangle \end{array}\)

Find \(\overrightarrow {PR} \).

\(\begin{array}{l}\overrightarrow {PR} = R(5,3,1) - P(0, - 2,0)\\\overrightarrow {PR} = \langle (5 - 0),(3 + 2),(1 - 0)\rangle \\\overrightarrow {PR} = \langle 5,5,1\rangle \end{array}\)

Re-Modify equation (1).

\(\begin{array}{l}\overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{l}}3&{ - 2}\\5&1\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}4&{ - 2}\\5&1\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}4&3\\5&5\end{array}} \right|{\rm{k}}\\\overrightarrow {PQ} \times \overrightarrow {PR} = (3 + 10){\rm{i}} - (4 + 10){\rm{j}} + (20 - 15){\rm{k}}\\\overrightarrow {PQ} \times \overrightarrow {PR} = 13{\rm{i - }}14{\rm{j + 5k}}\\\overrightarrow {PQ} \times \overrightarrow {PR} = \langle 13, - 14,5\rangle \end{array}\)

Thus, the nonzero vector orthogonal to the plane through the points \(P\), \(Q\) and \(R\) is\(\langle 13, - 14,5\rangle \).

calculating area of Triangle with the help of Formula.

(b)

Here, \({\rm{a}}\) and \({\bf{b}}\) are the vectors.

Re-modify equation\(\left( 2 \right)\).

Area of triangle \( = \frac{1}{2}|\overrightarrow {PQ} \times \overrightarrow {PR} |\)

We can Substituting, \(\langle 13, - 14,5\rangle \) for \(\overrightarrow {PQ} \times \overrightarrow {PR} \),

Area of triangle \( = \frac{1}{2}|\langle 13, - 14,5\rangle |\)

\(\begin{array}{l} = \frac{1}{2}\left( {\sqrt {{{(13)}^2} + {{( - 14)}^2} + {{(5)}^2}} } \right)\\ = \frac{1}{2}(\sqrt {169 + 196 + 25} )\\ = \frac{1}{2}(\sqrt {390} )\end{array}\)

Thus, the area of triangle \(P{\rm{ }}Q{\rm{ }}R\) is \(\frac{1}{2}(\sqrt {390} )\).