The equation of the line parallel to x-3 1 - y-3 5 - 2z-5 3 and passing through the point 1-3-5 in vector form- is-

The correct option is C r⃗ =( i⃗ +3j⃗ +5k⃗) +t( i⃗ +5j⃗ + 3 2 k⃗) We have to find the equation of the line parallel to x 31=y+35=2z 53 and passing ...

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Standard XII

Mathematics

Equation of a Plane : Vector Form

The equation ...

Question

The equation of the line parallel to $\frac{x - 3}{1} = \frac{y + 3}{5} = \frac{2 z - 5}{3}$ and passing through the point $(1, 3, 5)$ in vector form, is:

\to r = (\to i + 5 \to j + 3 \to k) + t (\to i + 3 \to j + 5 \to k)

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\to r = (\to i + 3 \to j + 5 \to k) + t (\to i + 5 \to j + 3 \to k)

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\to r = (\to i + 5 \to j + \frac{3}{2} \to k) + t (\to i + 3 \to j + 5 \to k)

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\to r = (\to i + 3 \to j + 5 \to k) + t (\to i + 5 \to j + \frac{3}{2} \to k)

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Solution

The correct option is C $\to r = (\to i + 3 \to j + 5 \to k) + t (\to i + 5 \to j + \frac{3}{2} \to k)$
We have to find the equation of the line parallel to $\frac{x - 3}{1} = \frac{y + 3}{5} = \frac{2 z - 5}{3}$ and passing through the point $(1, 3, 5)$ vector form.

We know that the equation of the line passing through the point with position vector $\to a$ and parallel to the vector $\to b$ is $\to r = \to a + t \to b$

Consider $\frac{x - 3}{1} = \frac{y + 3}{5} = \frac{2 z - 5}{3}$

Rewriting we get $\frac{x - 3}{1} = \frac{y + 3}{5} = \frac{z - \frac{5}{2}}{\frac{3}{2}}$

Thus vector representation is $\to b = \to i + 5 \to j + \frac{3}{2} \to k$

Since line passes through $(1, 3, 5)$ , $\to a = \to i + 3 \to j + 5 \to k$

Hence the required equation of the line is $\to r = (\to i + 3 \to j + 5 \to k) + t (\to i + 5 \to j + \frac{3}{2} \to k)$

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Similar questions

Q. The point of intersection of the lines

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

and

\to r = (- 2 \to i + 3 \to j + 5 \to k) + t (\to i + 2 \to j + 3 \to k)

Q. If

\to r = 3 \to i + 2 \to j - 5 \to k

\to a = 2 \to i - \to j + \to k

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meets the plane

\to r \cdot (2 \to i + 4 \to j - \to k) = 3

Q. The distance between the plane whose equation is

\to r \cdot (2 \to i + \to j - 3 \to k) = 5

and the line whose equation is

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,is

Q. Let

\to a = \to i + \to j - \to k, \to b = 5 \to i - 3 \to j - 3 \to k, \to c = 3 \to i - \to j + 2 \to k

If a vector

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is colinear with

\to c

and

| \to r | = ∣ ∣ \to a + \to b ∣ ∣

then

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equals.

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The equation of the line parallel to x−31=y+35=2z−53 and passing through the point (1,3,5) in vector form, is:

The equation of the line parallel to $\frac{x - 3}{1} = \frac{y + 3}{5} = \frac{2 z - 5}{3}$ and passing through the point $(1, 3, 5)$ in vector form, is: