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प्रश्न
Find the vector equation of the line passing through the point (–2, 1, 4) and perpendicular to the plane `barr*(4hati - 5hatj + 7hatk)` = 15
योग
उत्तर
Equation of given plane is
`barr*(4hati - 5hatj + 7hatk)` = 15
∴ The normal vector to the plane is
`barb = 4hati - 5hatj + 7hatk`
Since the required line is perpendicular to the given plane, the line is parallel to `barb`.
∴ The vector equation of the line passing
through `A(bara)` = (–2, 1, 4) is `barr = bara + λbarb`,
i.e. `barr = -2hati + hatj + 4hatk + λ(4hati - 5hatj + 7hatk)`
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