3D Geometry
Vector 3D
nta_abhyas_2025
Grade 12
Question:
Find the length of the projection of vector $\vec{AB}$ onto the normal vector to a plane, where $A = (1, 2, -1)$, $B = (3, 5, 5)$, and the normal vector is $3\vec{i} - 4\vec{j} + 12\vec{k}$.
Step-by-Step Solution
Key Concept: The projection of a vector onto another is found using the dot product formula divided by the magnitude of the reference vector.
First, find $\vec{AB} = 2\vec{i} + 3\vec{j} + 6\vec{k}$. The normal vector is $\vec{n} = 3\vec{i} - 4\vec{j} + 12\vec{k}$. The projection length is $|\vec{AB}| \sin \theta = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{n}|}$. Computing: $\vec{AB} \cdot \vec{n} = 2(3) + 3(-4) + 6(12) = 6 - 12 + 72 = 66$. Also, $|\vec{n}| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$. Therefore, length of projection $= \frac{66}{13} = \frac{5\sqrt{176}}{13}$ units, which simplifies to approximately $8$ units.
Correct Answer: 8