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NCERT Exemplar · Q18

Q.The vector equation of the line through the points (3,4,−7)(3, 4, -7) and (1,−1,6)(1, -1, 6) is __________.

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Appeared in past exams:MHT-CET 2021· Set pcm-2021-09-20-M· 2mreworded
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The vector equation of a line is r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}, where a⃗\vec{a} is a point on the line and b⃗\vec{b} is the direction vector. For points (3,4,−7)(3,4,-7) and (1,−1,6)(1,-1,6), the direction vector is (1−3,−1−4,6−(−7))=(−2,−5,13)(1-3, -1-4, 6-(-7)) = (-2, -5, 13). So the equation is r⃗=(3,4,−7)+λ(−2,−5,13)\vec{r} = (3,4,-7) + \lambda(-2,-5,13).

To write the vector equation of a line, you need two things: a position vector of any point on the line, and a direction vector that tells you which way the line runs. Think of it like this: you start at a fixed point (the position vector), and then you can move any distance along the direction vector — that sweep gives you every point on the line.

The general form is:

r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}

where a⃗\vec{a} is the position vector of a known point, b⃗\vec{b} is a direction vector parallel to the line, and λ\lambda is a scalar parameter (any real number).

Now let’s apply this to the given points.

  1. Pick a point on the line. You can choose either of the two given points. Let’s take (3,4,−7)(3, 4, -7). Its position vector is:

a⃗=3i^+4j^−7k^\vec{a} = 3\hat{i} + 4\hat{j} -7\hat{k}

(If you chose the other point, the final equation would look different but describe the same line — that’s fine.)

  1. Find the direction vector. The direction vector is simply the vector from one point to the other. Subtract the coordinates of the first point from the second (or vice versa — the direction just flips sign, which is still parallel). From (3,4,−7)(3,4,-7) to (1,−1,6)(1,-1,6):

b⃗=(1−3)i^+(−1−4)j^+(6−(−7))k^=−2i^−5j^+13k^\vec{b} = (1-3)\hat{i} + (-1-4)\hat{j} + (6-(-7))\hat{k} = -2\hat{i} -5\hat{j} + 13\hat{k}

So b⃗=(−2,−5,13)\vec{b} = (-2, -5, 13). …

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