Question

Find the angle between the lines whose direction cosines satisfy the equations
$l + m +n = 0$

${l}^{2} + {m} ^{2} - {n}^{2} = 0$


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Answer 1

$\large \red{\bf \: ⟼ Question :- } ✍$

Find the angle between the lines whose direction cosines satisfy the equations

$\bf \: l+ m+ n = 0 \: and \: {l}^{2} + {m}^{2} - {n}^{2} = 0$

$\huge \orange{AηsωeR}✍$

$\large \red{\bf \: ⟼ Given :- } ✍$

Two lines whose direction cosines satisy the equations

$\bf \: l+ m+ n = 0 \: and \: {l}^{2} + {m}^{2} - {n}^{2} = 0$

$\large \red{\bf \: ⟼ To \: Find :- } ✍$

• Angle between two lines

$\large \red{\bf \: ⟼ Formula Used :- } ✍$

Let us consider two lines havinh direction ratios (a, b, c) and (d, e, f) respectively and let 'x' be the angle between the lines, then

$\bf \:cosx = |\dfrac{a d + b e + cf}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } \sqrt{{d}^{2} + {e}^{2} + {f}^{2} } } |$

$\large \red{\bf \: ⟼ Solution :- } ✍$

Consider,

$\bf \: ⟼ \bf \: l+ m+ n = 0 \: ⟼ \: (1)$

$\bf \: ⟼ n = - l - m \: ⟼ \: (2)$

Consider,

$\bf \: {l}^{2} + {m}^{2} - {n}^{2} = 0$

On substituting the value of n from equation (2), we get

$\bf \: ⟼ {l}^{2} + {m}^{2} - {( - l - m)}^{2} = 0$

$\bf \: ⟼ {l}^{2} + {m}^{2} - {l}^{2} - {m}^{2} - 2mn = 0$

$\bf\implies \:2mn = 0$

$\bf\implies \:m = 0 \: ⟼(3) \: and \: n = 0 \: ⟼ \: (4)$

On solving equation (1) and (3), using cross multiplication method, we get

$\bf \: ⟼ l + m + n = 0$

$\bf \: ⟼ 0l + m + 0n = 0$

$\bf\implies \:\dfrac{l}{ - 1} = \dfrac{ - m}{0} = \dfrac{n}{1}$

$\bf\implies \:d. \: ratios \: of \: line \: (1) \: is \: (-1,0,1) \: ⟼ \: (5)$

On solving equation (1) and (4), using cross multiplication method, we get

$\bf \: ⟼ \bf \: l+ m+ n = 0$

$\bf \: ⟼ 0l + 0m + n = 0$

$\bf\implies \:\dfrac{l}{1} = \dfrac{m}{ - 1} = \dfrac{n}{0}$

$\bf\implies \:d. \: ratios \: of \: line \: (2) = (1,1,0) \: ⟼ \: (6)$

Let 'x' be the angle between the lines, then

$\bf \: ⟼ cosx \: = |\dfrac{ - 1 \times 1 + 0 \times 1 + 1 \times 0}{ \sqrt{1 + 0 + 1} \times \sqrt{1 + 1 + 0} }|$

$\bf\implies \:cosx = \dfrac{1}{2}$

$\bf \: ⟼ x = \dfrac{\pi}{3}$

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