Question 12 of 25beginner💡 UnderstandVery Short2 marks

The cartesian equation of a line is x−53=y+47=z−62\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}. Write its vector form.

Correct Answer

The given Cartesian equation of the line is x−53=y+47=z−62\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}. Comparing this with the standard form x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}, we identify the coordinates of the point as (5,−4,6)(5, -4, 6) and the direction ratios as 3,7,23, 7, 2.

Using these values, the position vector a⃗\vec{a} is 5i^−4j^+6k^5\hat{i} - 4\hat{j} + 6\hat{k} and the vector parallel to the line, b⃗\vec{b}, is 3i^+7j^+2k^3\hat{i} + 7\hat{j} + 2\hat{k}.

Therefore, the required vector form of the line is: r⃗=(5i^−4j^+6k^)+λ(3i^+7j^+2k^)\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k}).

Exercise: EXERCISE 11.2 | Q: 7 | (Page Page 13)
For More Understanding

Explanation

The context establishes the relationship between the Cartesian form (x−x1)/a=(y−y1)/b=(z−z1)/c(x-x_1)/a = (y-y_1)/b = (z-z_1)/c and the vector form r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}. By comparing the given equation with the standard Cartesian form, we extract the point coordinates (x1,y1,z1)(x_1, y_1, z_1) from the numerators and the direction ratios (a,b,c)(a, b, c) from the denominators. These are then converted into the position vector a⃗\vec{a} and the direction vector b⃗\vec{b} respectively to write the vector equation.

Solution Steps

  1. Step 1: Compare the given equation x−53=y+47=z−62\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} with the standard form x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} to find the point (x1,y1,z1)(x_1, y_1, z_1) and direction ratios (a,b,c)(a, b, c).

  2. Step 2: Identify the point as (5,−4,6)(5, -4, 6) since y+4y+4 is written as y−(−4)y-(-4), and the direction ratios as (3,7,2)(3, 7, 2).

  3. Step 3: Write the position vector a⃗=5i^−4j^+6k^\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k} and the direction vector b⃗=3i^+7j^+2k^\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}.

  4. Step 4: Substitute a⃗\vec{a} and b⃗\vec{b} into the vector equation formula r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b} to get r⃗=(5i^−4j^+6k^)+λ(3i^+7j^+2k^)\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k}).

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