Question 25 of 25advanced🔧 ApplyLong Answer6 marks

Find the vector equation of the line passing through the point (1,2,−4)(1, 2, -4) and perpendicular to the two lines: x−83=y+19−16=z−107\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} and x−153=y−298=z−5−5\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}

Correct Answer

r⃗=(i^+2j^−4k^)+λ(2i^+3j^+6k^)\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})

Exercise: Miscellaneous Exercise on Chapter 11 | Q: 5 | (Page Page 15)
For More Understanding

Explanation

To find the vector equation of a line, we need a point on the line and a direction vector. The point is given as (1,2,−4)(1, 2, -4), which corresponds to the position vector a⃗=i^+2j^−4k^\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}. The direction vector b⃗\vec{b} of the required line must be perpendicular to the direction vectors of the two given lines. We find b⃗\vec{b} by calculating the cross product of the direction vectors of the given lines. Finally, we substitute a⃗\vec{a} and b⃗\vec{b} into the standard vector equation form r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b} as demonstrated in Example 6 of the provided context.

Solution Steps

  1. Step 1: Identify position vector and direction vectors. The line passes through point (1,2,−4)(1, 2, -4), so its position vector is a⃗=i^+2j^−4k^\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}. The direction ratios of the first line are 3,−16,73, -16, 7, so its direction vector is b⃗1=3i^−16j^+7k^\vec{b}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}. The direction ratios of the second line are 3,8,−53, 8, -5, so its direction vector is b⃗2=3i^+8j^−5k^\vec{b}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k}.

  2. Step 2: Find the direction vector of the required line. The required line is perpendicular to both lines, so its direction vector b⃗\vec{b} is parallel to b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2.

    b⃗=∣i^j^k^3−16738−5∣\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} b⃗=i^((−16)(−5)−(7)(8))−j^((3)(−5)−(7)(3))+k^((3)(8)−(−16)(3))\vec{b} = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3)) b⃗=i^(80−56)−j^(−15−21)+k^(24+48)\vec{b} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) b⃗=24i^+36j^+72k^\vec{b} = 24\hat{i} + 36\hat{j} + 72\hat{k}

    Simplifying by dividing by 12, we get b⃗=2i^+3j^+6k^\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}.

  3. Step 3: Write the vector equation. Using the form r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b} (as shown in Example 6), we substitute the values: r⃗=(i^+2j^−4k^)+λ(2i^+3j^+6k^)\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}).

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