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The vector equation of the line passing through points $A(3,4,-7)$ and $B(1,-1,6)$ is |
$\vec x=3\hat i+4\hat j-7\hat k+λ(\hat i-\hat j+6\hat k)$ $\vec x=(3-2λ)\hat i+ (4 − 5λ)\hat j + (-7 +13λ)\hat k$ $\vec x =\hat i−\hat j+6\hat k+ λ(3\hat i+4\hat j − 7\hat k)$ $\vec x = (−2+3λ)\hat i+ (−5 +4λ)\hat j + (13 – 7λ)\hat k$ |
$\vec x=(3-2λ)\hat i+ (4 − 5λ)\hat j + (-7 +13λ)\hat k$ |
The correct answer is Option (2) → $\vec x=(3-2λ)\hat i+ (4 − 5λ)\hat j + (-7 +13λ)\hat k$ Points are $A(3,4,-7)$ and $B(1,-1,6)$. Direction vector $\vec{AB}=(1-3,\,-1-4,\,6-(-7))=(-2,-5,13)$. Vector equation of the line through $A$ in direction $\vec{AB}$: $\vec x=\vec a+\lambda\vec{AB}$ $\vec x=(3,4,-7)+\lambda(-2,-5,13)$ $\vec x=(3-2\lambda)\hat i+(4-5\lambda)\hat j+(-7+13\lambda)\hat k$ Final answer: $\vec x=(3-2\lambda)\hat i+(4-5\lambda)\hat j+(-7+13\lambda)\hat k$ |
