Equation of line of shortest distance between the skew lines- r -3 -5 -7 k-2 -k-r -k-7 -6 -k-A- r -3 i -5 j -7 k - t 2 i -3 j -4 k B- r -3 i -5 j -7 k - t 2 i -3 j -4 k C- r -3 i -5 j -7 k - t 2 i -3 j -4 k D- r -3 i -5 j -7 k - t 2 i -3 j -4 k

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Equation of a Plane : Point Normal Form

Equation of l...

Question

Equation of line of shortest distance between the skew lines. $¯ r = (3^i + 5^j + 7^k) + λ (^i - 2^j +^k)$
$¯ r = (-^i -^j -^k) + μ (7^i - 6^j +^k)$

¯ r = (3 i + 5 j + 7 k) + t (2 i + 3 j + 4 k)

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¯ r = (3 i - 5 j + 7 k) + t (2 i + 3 j + 4 k)

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¯ r = (- 3 i + 5 j - 7 k) + t (2 i + 3 j + 4 k)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

¯ r = (3 i - 5 j - 7 k) + t (2 i + 3 j + 4 k)

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Solution

The correct option is A $¯ r = (3 i + 5 j + 7 k) + t (2 i + 3 j + 4 k)$

$¯ P Q = (P . V o f Q) - (P . V . o f P)$
$= L_{2} - L_{1}$
$= (7 μ - λ - 4) i + (- 6 μ + 2 λ - 6) j) + (μ - λ - 8) K$
$¯ P Q$ is perpendicular to $L_{1} \Rightarrow (P Q) . L_{1} = 0$
$\Rightarrow 10 μ - 3 λ = 0 \to (1)$
$¯ P Q$ is perpendicular to $L_{2} = (P Q) . L_{2} = 0$
$=\Rightarrow 43 μ - 10 λ = 0 \to (1)$
On solving (1) & (2) we get $λ = 0, μ = 0 \Rightarrow P Q = - 4 i - 6 j - 8 k$
Since PQ passes through P and parallel to $¯ P Q$ we have
$¯ r = (3 i + 5 j + 7 k) + 0 (i - 2 j + k) + λ (- 4 i - 6 j - 8 k)$
$= (3 i + 5 j + 7 k) + t (2 i + 3 j + 4 k)$ where $t = - 2 λ$

Suggest Corrections

Similar questions

Q. Find the shortest distance between the following pairs of lines whose vector equations are:

(i)

\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + λ (3 \hat{i} - \hat{j} + \hat{k}) and \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})

(ii)

\vec{r} = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) + λ (\hat{i} - 2 \hat{j} + 7 \hat{k}) and \vec{r} = - \hat{i} - \hat{j} - \hat{k} + μ (7 \hat{i} - 6 \hat{j} + \hat{k})

(iii)

\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) and \vec{r} = (2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + μ (3 \hat{i} + 4 \hat{j} + 5 \hat{k})

(iv)

\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - t) \hat{k} and \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}

(v)

\vec{r} = (λ - 1) \hat{i} + (λ + 1) \hat{j} - (1 + λ) \hat{k} and \vec{r} = (1 - μ) \hat{i} + (2 μ - 1) \hat{j} + (μ + 2) \hat{k}

(vi)

\vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + λ (2 \hat{i} - 5 \hat{j} + 2 \hat{k}) and, \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})

(vii)

\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k}) and, \vec{r} = 2 \hat{i} + \hat{j} - \hat{k} + μ (3 \hat{i} - 5 \hat{j} + 2 \hat{k})

(viii)

\vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k}

and

\vec{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})

[NCERT EXEMPLAR]

Q. Show that the line whose vector equation is

\vec{r} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k} + λ (\hat{i} + 3 \hat{j} + 4 \hat{k})

is parallel to the plane whose vector equation is

\vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 7 .

Also, find the distance between them.

The shortest distance between the lines $\to r = 3 \to i + 5 \to j + 7 \to k + λ (\to i + 2 \to j + \to k) a n d \to r = - \to i - \to j - \to k + μ (7 \to i - 6 \to j + \to k) i s$

Q. Find the vector equation of the plane passing through points

4 i - 3 j - k, 3 i + 7 j - 10 k

and

2 i + 5 j - 7 k

and show that the point

i + 2 j - 3 k

lies in the plane.

Q. Test whether the lines

¯ ¯ ¯ p = (¯ ¯¯¯ ¯ 2 i + ¯ ¯¯¯ ¯ 4 j + ¯ ¯¯¯¯ ¯ 6 k) + λ (¯ i + ¯ ¯¯¯ ¯ 4 j + ¯ ¯¯¯¯ ¯ 7 k)

and

q = (¯ ¯¯¯¯ ¯ - i - ¯ ¯¯¯ ¯ 3 j - ¯ ¯¯¯¯ ¯ 5 k) + ω (¯ ¯¯¯ ¯ 3 i + ¯ ¯¯¯ ¯ 5 j + ¯ ¯¯¯¯ ¯ 7 k)

are co-planer .if so find the equation of the plane containing them.

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Equation of line of shortest distance between the skew lines. ¯r=(3^i+5^j+7^k)+λ(^i−2^j+^k) ¯r=(−^i−^j−^k)+μ(7^i−6^j+^k)

Equation of line of shortest distance between the skew lines. $¯ r = (3^i + 5^j + 7^k) + λ (^i - 2^j +^k)$
$¯ r = (-^i -^j -^k) + μ (7^i - 6^j +^k)$