Questions
The vector equation of the plane through the point and to the line of intersection of the plane and is
detailed solution
Correct option is B
The line of intersection of the planes r→.(3i^−j^+k^)=1 and r→.(i^+4j^−2k^)=2 is common to both the planes. Therefore, it is ⊥ to normals to the two planes i.e. n→1=3i^−j^+k^ and n→2=i^+4j^−2k^ . Hence it is parallel to the vector n→1×n→2=−2i^+7j^+13k^ . Thus, we have to find the equation of the plane passing through a→=i^+2j^−k^ and normal to the vector n→=n1→×n2→ . The equation of the required plane is(r→−a→).n→ or r→.n→=a→.n→ or r→.(−2j^+7j^+13k^)=(i^+2j^−k^).(−2j^+7j^+13k^) or r→.(2i^−7j^−13k^)=1Similar Questions
If the vectors are mutually perpendicular and if are also mutually perpendicular, then the cosine of the angle between is
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