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Find the coordinates of the point where the line through (5,1,6) and `3hati+4hatj+hatk` crosses the yz-plane.

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To find the coordinates of the point where the line through the points (5, 1, 6) and (3, 4, 1) crosses the yz-plane, we can follow these steps: ### Step 1: Define the Points Let the two points be: - Point A: \( A(3, 4, 1) \) which corresponds to the vector \( \vec{A} = 3\hat{i} + 4\hat{j} + 1\hat{k} \) - Point B: \( B(5, 1, 6) \) which corresponds to the vector \( \vec{B} = 5\hat{i} + 1\hat{j} + 6\hat{k} \) ### Step 2: Find the Direction Vector The direction vector \( \vec{B} - \vec{A} \) can be calculated as follows: \[ \vec{B} - \vec{A} = (5 - 3)\hat{i} + (1 - 4)\hat{j} + (6 - 1)\hat{k} = 2\hat{i} - 3\hat{j} + 5\hat{k} \] ### Step 3: Write the Parametric Equation of the Line The equation of the line can be expressed in parametric form as: \[ \vec{R} = \vec{A} + \lambda(\vec{B} - \vec{A}) \] Substituting the values: \[ \vec{R} = (3\hat{i} + 4\hat{j} + 1\hat{k}) + \lambda(2\hat{i} - 3\hat{j} + 5\hat{k}) \] This expands to: \[ \vec{R} = (3 + 2\lambda)\hat{i} + (4 - 3\lambda)\hat{j} + (1 + 5\lambda)\hat{k} \] ### Step 4: Find the Intersection with the yz-plane The yz-plane is defined by the condition \( x = 0 \). Therefore, we set: \[ 3 + 2\lambda = 0 \] Solving for \( \lambda \): \[ 2\lambda = -3 \implies \lambda = -\frac{3}{2} \] ### Step 5: Substitute \( \lambda \) Back into the Line Equation Now substitute \( \lambda = -\frac{3}{2} \) into the parametric equations for \( y \) and \( z \): - For \( y \): \[ y = 4 - 3\left(-\frac{3}{2}\right) = 4 + \frac{9}{2} = \frac{8}{2} + \frac{9}{2} = \frac{17}{2} \] - For \( z \): \[ z = 1 + 5\left(-\frac{3}{2}\right) = 1 - \frac{15}{2} = \frac{2}{2} - \frac{15}{2} = -\frac{13}{2} \] ### Step 6: Write the Final Coordinates Thus, the coordinates of the point where the line crosses the yz-plane are: \[ (0, \frac{17}{2}, -\frac{13}{2}) \] ### Summary of the Solution The coordinates of the intersection point are: \[ \boxed{(0, \frac{17}{2}, -\frac{13}{2})} \]
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