Question

show that the line joining the points A(2,3,-1) and B(-3,5,-3) Is perpendicular the line joining C(1,2,3)and D(3,5,7)​

Answer 1

Question:

Show that the line joining the points A (2, 3, -1) and B (3, 5, -3) is perpendicular to the line joining C (1, 2, 3) and (3, 5, 7).

Answer:

Step-by-step explanation:

Given:

• A line joining the points A (2, 3, -1) and B (3, 5, -3)
• A line joining the points C (1, 2, 3) and D (3, 5, 7)

To Prove:

• The lines are perpendicular to each other

Proof:

For two lines to be perpendicular, the dot product of their direction vectors are 0.

That is,

l₁l₂ + m₁m₂ + n₁n₂ = 0

First find the direction ratios of the lines AB and CD

Drs of AB = (3-2, 5-3, -3+1)

= (1, 2, -2)

Drs of CD = (3-1, 5-2, 7-3)

= (2, 3, 4)

Now finding the direction cosines(l₁, m₁, n₁) of line AB

We know that

$\tt l= \pm \dfrac{a}{\sqrt{a^2+b^2+c^2} } ,m=\pm \dfrac{b}{\sqrt{a^2+b^2+c^2} },n= \pm \dfrac{c}{\sqrt{a^2+b^2+c^2} }$

Hence,

$\tt Dcs\:of\:line\:AB=\dfrac{1}{\sqrt{1+4+4} } ,\dfrac{2}{\sqrt{1+4+4} } ,\dfrac{-2}{\sqrt{1+4+4} }$

$\tt Dcs\:of\:line\:AB=\dfrac{1}{3 } ,\dfrac{2}{3 } ,\dfrac{-2}{3 }--(1)$

Now finding direction cosines (l₂, m₂, n₂) of line CD

Hence,

$\tt Dcs\:of\:line\:CD=\dfrac{2}{\sqrt{4+9+16} } ,\dfrac{3}{\sqrt{4+9+16} } ,\dfrac{4}{\sqrt{4+9+16} }$

$\tt Dcs\:of\:line\:CD=\dfrac{2}{\sqrt{29} } ,\dfrac{3}{\sqrt{29} } ,\dfrac{4}{\sqrt{29} }---(2)$

From 1 and 2,

$\tt \dfrac{2}{\sqrt{29} } \times \dfrac{1}{3 } +\dfrac{3}{\sqrt{29} }\times\dfrac{2}{3 } +\dfrac{4}{\sqrt{29} }\times \dfrac{-2}{3 }$

$\implies \tt\dfrac{2}{3\sqrt{29} } +\dfrac{6}{3\sqrt{29} } +\dfrac{-8}{3\sqrt{29} }$

$\implies \tt +\dfrac{8}{3\sqrt{29} } +\dfrac{-8}{3\sqrt{29} }$

= 0

Hence the lines are perpendicular to each other.

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

Show that the line joining the points A(2,3,-1) and B(-3,5,-3) Is perpendicular the line joining C(1,2,3)and D(3,5,7)

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$\huge {AηsωeR} ✍$

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$\begin{gathered}\begin{gathered}\bf Given = \begin{cases} &\sf{Line \: joining \: the \: points \: A(2,3,-1) \: and \: B(-3,5,-3).} \\ &\sf{Line \: joining \: the \: points \: C(1,2,3) \: and \: D(3,5,7).} \end{cases}\end{gathered}\end{gathered}$

To Show :-

• Line Segment AB is perpendicular to CD.

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Concept Used :-

☆Let us consider two lines having direction ratios

$\sf \:  (a_1,b_1,c_1) \: and \: (a_2,b_2,c_2) \:$

$\sf \:  then \: lines \: are \: perpendicular \: iff$

$\bf \:  a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$

☆Let us consider two points A and B having coordinates

$\sf \:  ⟼(x_1,y_1,z_1) \: and \: (x_2,y_2,z_2) \: then \: direction \: ratio \: is$

$\bf \:  ⟼(x_2 - x_1,y_2 - y_1,z_2 - z_1)$

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$\large\underline\purple{\bold{Solution :- }}$

☆Let us take first Line segment joining the points A(2,3,-1) and B(-3,5,-3).

☆So direction ratios of line segment AB is given by

$\sf \:  ⟼d.r's \: of \: AB = ( - 3 - 2 ,5 - 3 , - 3 + 1)$

$\sf \:  ⟼d.r's \: of \: AB = ( - 5 ,2 , - 2)$

$\bf\implies \:a_1 = - 5,b_1 = 2,c_1 = - 2$

☆Let us take second Line segment joining the points C(1,2,3)and D(3,5,7).

☆So, direction ratios of line segment CD is given by

$\sf \:  ⟼d.r's of \: CD \: = (3 - 1 ,5 - 2 ,7 - 3)$

$\sf \:  ⟼d.r's \: of \: CD = (2 ,3 ,4)$

$\bf\implies \:a_2 = 2, \: b_2 = 3 \: ,c_2 \: = 4$

☆ To show that Line segment AB is perpendicular to Line segment CD,

☆ Let us assume

$\sf \:  ⟼a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}$

$\sf \:  ⟼ - 5 \times 2 + 3 \times 2 + ( - 2) \times 4$

$\sf \:  ⟼ - 10 + 6 - 8 = - 12 \: ≠ \: 0$

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Hence, Line segment AB is not perpendicular to CD.