<p><strong>39. If <math><mrow><msub><mrow><mi>l</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo> </mo><msub><mrow><mi>m</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo> </mo><msub><mrow><mi>n</mi></mrow><mrow><mn>1</mn></mrow></msub><mo> </mo><mi>a</mi><mi>n</mi><mi>d</mi><mo> </mo><msub><mrow><mi>l</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo> </mo><msub><mrow><mi>m</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>,</mo><mo> </mo><msub><mrow><mi>n</mi></mrow><mrow><mn>2</mn></mrow></msub><mo> </mo></mrow></math> are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are <math><mrow><msub><mrow><mi>m</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>n</mi></mrow><mrow><mn>2</mn></mrow></msub><mo> </mo><mo>−</mo><mo> </mo><msub><mrow><mi>m</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>n</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo> </mo><msub><mrow><mi>n</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>l</mi></mrow><mrow><mn>2</mn></mrow></msub><mo> </mo><mo>−</mo><mo> </mo><msub><mrow><mi>n</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>l</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><mo> </mo><msub><mrow><mi>l</mi></mrow><mrow><mn>1</mn></mrow></msub><msub><mrow><mi>m</mi></mrow><mrow><mn>2</mn></mrow></msub><mo> </mo><mo></mo><mo>−</mo><mo> </mo><msub><mrow><mi>l</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi>m</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>.</mo></mrow></math></strong></p>

It is given that l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines. Therefore,l1l2+m1m2+n1 n2=0..........(1)l12+m12+n1 2=1..........(2)l22+m22+n2 2=1..........(3)Let l, m, n be the direction cosines of the line which is perpendicular to the line with direction cosines l1, m1, n1 and l2, m2, n2.∴ll1+ mm1+ nn1 =0l l2+m m2+n n2=0∴lm1n2−m2n1=mn1l2−n2l1=nl1m2−l2m1⇒l2(m1n2−m2n1)2=m2(n1l2−n2l1)2=n2(l1m2−l2m1)2⇒l2(m1n2−m2n1)2=m2(n1l2−n2l1)2=n2(l1m2−l2m2)2=l2+m2+n2(m1n2−m2n1)2+(n1l2−n2l1)2+(l1m2−l2m2)2..........(4)l, m, n are the direction cosines of the line.∴l2 + m2 + n2 =1…(5)It is known that,(l12+m12+n1 2)(l22+m22+n2 2)−(l1l2+m1m2+n1 n2)2=(m1n2−m2n1)2+(n1l2−n2l1)2+(l1m2−l2m1)2From,(1),(2)&(3),we.obtain⇒1.1−0=(m1n2−m2n1)2+(n1l2−n2l1)2+(l1m2−l2m1)2∴(m1n2−m2n1)2+(n1l2−n2l1)2+(l1m2−l2m1)2=1..........(6)Substituting the values from equations (5) and (6) in equation (4), we obtainl2(m1n2−m2n1)2=m2(n1l2−n2l1)2=n2(l1m2−l2m1)2=1Thus, the direction cosines of the required line are m1n2−m2n1,n1l2−n2l1,l1m2−l2m1

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Ncert Solutions Maths class 12th Maths Class 12th Three Dimensional Geometry

39. If $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1} .$

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Vishal Baghel | Contributor-Level 10

4 months ago

It is given that $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines. Therefore,

$\begin{array}{l} l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 . . . . . . . . . . (1) \\ {l_{1}}^{2} + {m_{1}}^{2} + n_{1}^{2} = 1 . . . . . . . . . . (2) \\ {l_{2}}^{2} + {m_{2}}^{2} + n_{2}^{2} = 1 . . . . . . . . . . (3) \end{array}$

Let $l, m, n$ be the direction cosines of the line which is perpendicular to the line with direction cosines $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2} .$

$\begin{array}{l} ∴ l l_{1} + m m_{1} + n n_{1} = 0 \\ l l_{2} + m m_{2} + n n_{2} = 0 \\ ∴ \frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{n_{1} l_{2} - n_{2} l_{1}} = \frac{n}{l_{1} m_{2} - l_{2} m_{1}} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{1})}^{2}} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{2})}^{2}} \\ = \frac{l^{2} + m^{2} + n^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{2})}^{2}} . . . . . . . . . . (4) \end{array}$

$l, m, n$ are the direction cosines of the line.

$∴ l^{2} + m^{2} + n^{2} = 1 \dots (5)$

It is known that,

$\begin{array}{l} ({l_{1}}^{2} + {m_{1}}^{2} + n_{1}^{2}) ({l_{2}}^{2} + {m_{2}}^{2} + n_{2}^{2}) - {(l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})}^{2} \\ = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} \\ F r o m, (1), (2) & (3), w e . o b t a i n \\ \Rightarrow 1.1 - 0 = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} \\ ∴ {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} = 1 . . . . . . . . . . (6) \end{array}$

Substituting the values from equations (5) and (6) in equation (4), we obtain

$\frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{1})}^{2}} = 1$

Thus, the direction cosines of the required line are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1}$

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If $\int_{}^{} (\frac{x^{2} c o s x}{1 + π^{x}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) d x = \frac{π}{4} (π + α) - 2$

Then the value of 'α' is equal to

alok kumar singh

$\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{x^{2} c o s x}{1 + π^{2}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) d x = \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} {(\frac{x^{2} c o s x}{1 + π^{x}} + \frac{1 + s i n^{2} x}{1 + e^{{(s i n x)}^{2023}}}) + (\frac{x^{2} c o s x}{1 + π^{- x}} + \frac{1 + s i n^{2} x}{1 + e^{- {(s i n x)}^{2023}}})} d x$

$= \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} (x^{2} c o s x + 1 + s i n^{2} x) d x = \frac{π}{4} (π + α) - 2$

$\int_{0}^{\frac{π}{2}} x^{2} c o s x d x + \int_{0}^{\frac{π}{2}} (1 + s i n^{2} x) d x = \frac{π}{4} (π + α) - 2$ ....(1)

Let $I_{1} = \int_{0}^{\frac{π}{2}} (1 + s i n^{2} x) d x$

$I_{1} = \int_{0}^{\frac{π}{2}} 1 \cdot d x + \int_{0}^{\frac{π}{2}} (\frac{1 - c o s 2 x}{2}) d x$

$I_{1} = \frac{π}{2} + \frac{1}{2} [\frac{π}{2} + 0]$

$I_{1} = \frac{3 π}{4}$

Let $I_{2} = \int_{0}^{\frac{π}{2}} x^{2} c o s x d x$

$I_{2} = {[x^{2} (s i n x) - \int 2 x \int c o s x d x]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} (s i n x) - 2 \int x s i n x]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} s i n x - 2 (x (- c o s x) + \int c o s x)]}_{0}^{\frac{π}{2}}$

$I_{2} = {[x^{2} s i n x - 2 (- x c o s x + s i n x)]}_{0}^{\frac{π}{2}}$

$I_{2} = (\frac{π^{2}}{4} - 2)$

Put l₁ and l₂ in (1)

$\frac{π^{2}}{4} - 2 + \frac{3 π}{4}$

$\frac{π^{2}}{4} + \frac{3 π}{4} - 2$

$\frac{π}{4} (π + 3) - 2$

α = 3

If $| \vec{a} | = 1$ , $| \vec{b} | = 4$ $\vec{a} \cdot \vec{b} = 2$ and $\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$ Then the angle between $\vec{b}$ and $\vec{c}$ is

alok kumar singh

Given $| \vec{a} | = 1$ , $| \vec{b} | = 4$ , $\vec{a} \cdot \vec{b} = 2$

$\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$

Dot product with $\vec{a}$ on both sides

$\vec{c} \cdot \vec{a} = - 6$ ... (1)

Dot product with $\vec{b}$ on both sides

$\vec{b} \cdot \vec{c} = - 48$ ... (2)

$\vec{c} \cdot \vec{c} = 4 {| \vec{a} \times \vec{b} |}^{2} + 9 {| \vec{b} |}^{2}$

${| \vec{c} |}^{2} = 4 [{| \vec{a} |}^{2} \cdot {| \vec{b} |}^{2} - {(\vec{a} \cdot \vec{b})}^{2}] + 9 {| \vec{b} |}^{2}$

${| \vec{c} |}^{2} = 4 [(1) {(4)}^{2} - (4)] + 9 (16)$

${| \vec{c} |}^{2} = 4 [12] + 144$

${| \vec{c} |}^{2} = 48 + 144$

${| \vec{c} |}^{2} = 192$

$c o s θ = \frac{\vec{b} \cdot \vec{c}}{| \vec{b} | | \vec{c} |}$

$c o s θ = \frac{- 48}{\sqrt[]{192} \cdot 4}$

$c o s θ = \frac{- 48}{8 \sqrt[]{3} \cdot 4}$

$c o s θ = \frac{- 3}{2 \sqrt[]{3}}$

$c o s θ = \frac{- \sqrt[]{3}}{2}$ $θ = c o s^{- 1} (\frac{- \sqrt[]{3}}{2})$

If the foot of perpendicular from (1, 2, 3) to the line $\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z - 4}{1}$ is (a, b, g) then find a + b + g

alok kumar singh

(a – 1) × 2 + (b – 2) × 5 + (g – 3) × 1 = 0

2a + 5b + g – 15 = 0

Also, P lie on line

a + 1 = 2λ

b – 2 = 5λ

g – 4 = λ

2 (2λ – 1) + 5 (5λ + 2) + λ + 4 – 15 = 0

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

$λ = \frac{1}{10}$

a + b + g = (2λ – 1) + (5λ + 2) + (λ + 4)

$= 8 λ + 5 = \frac{8}{10} + 5 = 5.8$

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alok kumar singh

Take $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

$\vec{A B}$ = (α − 2) $\hat{i}$ + (β − 3) $\hat{j}$ + (γ − 4) $\hat{k}$

Now,

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2α − 4 + 3β − 9 + 4γ −16 = 0

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Let $λ$ be an integer. IF the shortest distance between the lines $x - λ = 2 y - 1 = - 2 z$ and $x = y + 2 λ = z - λ i s \frac{\sqrt[]{7}}{2 \sqrt[]{2}}$ , then the value of $| λ | i s . . . . . . . . . . . . . . . . . . .$

Vishal Baghel

$L_{1} = \frac{x - λ}{1} = \frac{y - \frac{1}{2}}{\frac{1}{2}} = \frac{z}{- \frac{1}{2}}$

$∴ S D = \frac{| - 2 λ + 3 (2 λ + \frac{1}{2}) + λ |}{\sqrt[]{14}} = \frac{| 5 λ + \frac{3}{2} |}{\sqrt[]{14}}$

$5 λ + \frac{3}{2} = - \frac{7}{2} \Rightarrow 5 λ = - 5 \Rightarrow λ = - 1$

$∴ | λ | = 1$

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39. If $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1} .$

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39. If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m1n2 − m2n1, n1l2 − n2l1, l1m2 ­− l2m1.

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39. If $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1} .$