If three mutually perpendicular lines have direction cosines - 1 - m 1 - n 1 - - 2 - m 2 - n 2 and - 3 - m 3 - n 3 - then the line having direction cosines - 1 - - 2 - - 3 - m 1 - m 2 - m 3 and n 1 - n 2 - n 3 make an angle of - with each other-

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

If three mutu...

Question

If three mutually perpendicular lines have direction cosines $(ℓ_{1}, m_{1}, n_{1}), (ℓ_{2}, m_{2}, n_{2})$ and $(ℓ_{3}, m_{3}, n_{3}),$ then the line having direction cosines $ℓ_{1} + ℓ_{2} + ℓ_{3}, m_{1} + m_{2} + m_{3}$ and $n_{1} + n_{2} + n_{3}$ make an angle of $\dots \dots$ with each other.

0^{\circ}

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30^{\circ}

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60^{\circ}

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90^{\circ}

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Solution

The correct option is A $0^{\circ}$
Let direction vector of three mutually perpendicular lines be

$\to a = l_{1}^i + m_{1}^j + n_{1}^k$

$\to b = l_{2}^i + m_{2}^j + n_{2}^k$

$\to c = l_{3}^i + m_{3}^j + n_{3}^k$

Let the direction vector associated with directions cosine $l_{1} + l_{2} + l_{3}, m_{1} + m_{2} + m_{3}, n_{1} + n_{2} + n_{3}$ be
$\to p = (l_{1} + l_{2} + l_{3})^i + (m_{1} + m_{2} + m_{3})^j + (n_{1} + n_{2} + n_{3})^k$

As, lines associated with direction vectors $a, b$ and $c$ are mutually perpendicular, we have

$\to a . \to b = 0$ [dot product of two perpendicular vector is $0$ ]

$\Rightarrow l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0$ ........ $(1)$

$\to a . \to c = 0$ [dot product of two perpendicular vector is $0$ ]

$\Rightarrow l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3} = 0$ ........ $(2)$

And $\to b . \to c = 0$ [dot product of two perpendicular vector is $0$ ]

$\Rightarrow l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} = 0$ ........ $(3)$

Now, let $x, y, z$ be the angles made by direction vectors $a, b$ and $c$ respectively with $p$

$∴$ ,

$cos x = \to a . \to p$

$\Rightarrow cos x = l_{1} (l_{1} + l_{2} + l_{3}) + m_{1} (m_{1} + m_{2} + m_{3}) + n_{1} (n_{1} + n_{2} + n_{3})$

$\Rightarrow cos x = {l_{1}}^{2} + l_{1} l_{2} + l_{1} l_{3} + {m_{1}}^{2} + m_{1} m_{2} + m_{1} m_{3} + {n_{1}}^{2} + n_{1} n_{2} + n_{1} n_{3}$

$\Rightarrow cos x = ({l_{1}}^{2} + {m_{1}}^{2} + {n_{1}}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}) + (l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3})$

As we know ${l_{1}}^{2} + {m_{1}}^{2} + {n_{1}}^{2} = 1 ∵$ sum of squares of direction cosines of a line is equal to $1$

$\Rightarrow cos x = 1 + 0 = 1$ from $(1)$ and $(2)$

$cos y = \to b . \to p$

$\Rightarrow cos y = l_{2} (l_{1} + l_{2} + l_{3}) + m_{2} (m_{1} + m_{2} + m_{3}) + n_{2} (n_{1} + n_{2} + n_{3})$

$\Rightarrow cos y = l_{2} l_{1} + {l_{2}}^{2} + l_{2} l_{3} + m_{2} m_{1} + {m_{2}}^{2} + m_{2} m_{3} + n_{2} n_{1} + {n_{2}}^{2} + n_{2} n_{3}$

$\Rightarrow cos y = ({l_{2}}^{2} + {m_{2}}^{2} + {n_{2}}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}) + (l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3})$

As we know ${l_{2}}^{2} + {m_{2}}^{2} + {n_{2}}^{2} = 1 ∵$ sum of squares of direction cosines of a line is equal to $1$

$\Rightarrow cos y = 1 + 0 = 1$ from $(1)$ and $(3)$

$cos z = \to c . \to p$

$\Rightarrow cos z = l_{3} (l_{1} + l_{2} + l_{3}) + m_{3} (m_{1} + m_{2} + m_{3}) + n_{3} (n_{1} + n_{2} + n_{3})$

$\Rightarrow cos z = l_{3} l_{1} + l_{3} l_{2} + {l_{3}}^{2} + m_{3} m_{1} + m_{3} m_{2} + {m_{3}}^{2} + n_{3} n_{1} + n_{3} n_{2} + {n_{3}}^{2}$

$\Rightarrow cos z = ({l_{3}}^{2} + {m_{3}}^{2} + {n_{3}}^{2}) + (l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3}) + (l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3})$

As we know ${l_{3}}^{2} + {m_{3}}^{2} + {n_{3}}^{2} = 1 ∵$ sum of squares of direction cosines of a line is equal to $1$

$\Rightarrow cos z = 1 + 0 = 1$ from $(2)$ and $(3)$

$∴ cos x = cos y = cos z = 1$

$\Rightarrow x = y = z = 0$

Hence, angle made by vector $p$ , with vectors $a, b$ and $c$ are equal.

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