MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Multiplication with Vectors

Find the shor...

Question

Find the shortest distance between the following pair of lines.
$¯ r = (^i + 2^j +^k) + λ (2^i -^j + 3^k)$
$¯ r = (^i - 2^j +^k) + μ (3^i + 2^j - 5^k)$

Open in App

Solution

Given two lines, $_{1} = (^i + 2^j +^k) + λ (2^i -^j + 3^k)$
$_{2} = (^i - 2^j +^k) + μ (3^i + 2^j - 5^k)$
The shortest distance between $r_{1}$ and $r_{2}$ must be perpendicular to both
$∴$ Distance is along $(2^i -^j + 3^k) \times (3^i + 2^j - 5^k)$
$∣ ∣ ∣ ∣ \begin{matrix} i & j & k 2 & - 1 & 3 3 & 2 & - 5 \end{matrix} ∣ ∣ ∣ ∣ = 0 i + 19 j + 1 k$
$= 19 j + 7 k$
$∴$ Shortest distance $= | [(i + 2 j + k) - (i - 3 j - k)] (19 j + 7 k) |$
$= | (5 j + 2 k) (19 j + 7 k) |$
$= 109$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Find shortest distance between lines

\to r =^i + 2^j + 3^k + λ (2^i +^j + 4^k)

$\to r = 2^i + 4^j + 5^k + μ (3^i + 4^j + 5^k)$

Q. Consider the following three lines in space

L_{1} : ¯ r = 3^i -^j + 2^k + λ (2^i + 4^j -^k);

L_{2} : ¯ r =^i +^j - 3^k + μ (4^i + 2^j + 4^k);

L_{3} : ¯ r = 3^i - 2^j - 2^k + t (2^i + 4^j - 2^k)

Which one of the following pair(s) are in the same plane

Q. Angle between the lines

¯ r = (^i + 2^j -^k) + μ (^i -^j +^k)

and Normal to the plane

¯ r = (2^i -^j +^k)

Q. Find the shortest distance between the following pairs of parallel lines.

\to r = (^i + 2^j + 3^k) + λ (^i -^j +^k)

\to r = (2^i -^j -^k) + μ (-^i +^j -^k)

Q. Find the vector

¯ r

perpendicular to

¯ a =^i - 2^j + 5^k

¯ b = 2^i + 3^j -^k

¯ r . (2^i +^j +^k) + 8 = 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Basic Operations

PHYSICS

Watch in App

explore_more_icon

Explore more

Multiplication with Vectors

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon