
The equation of the plane perpendicular to the yz-plane and passing through the point (1, -2, 4) and (3, -4, 5) is,
A.
B.
C.
D.
Answer
541.5k+ views
Hint: Plane perpendicular to the yz-plane is also parallel to the x-axis. So find the equation of the vector lying in the plane and the vector representing the normal vector. Thus take their cross product and find the equation of the plane using the point and normal.
Complete step-by-step answer:
It is said that a plane is perpendicular to the yz-plane, then the plane will be parallel to the x-axis. If the planes are perpendicular, then their normals are also perpendicular. Thus the normal vector, can be said as (1, 0, 0), as the normal of the x-axis, parallel to plane and yz are perpendicular.
Now we need to find the vector which is lying in the plane by using the 2 given points (1, -2, 4) and (3, -4, 5). Let us mark that vector as
Similarly, the parallel vector can be represented from point (1, 0, 0) as
So the normal of the plane is given as,
Let us take the cross product.
Therefore the equation of plane using the point (1, -2, 4) and normal is,
Hence we got the equation of plane perpendicular to the yz-plane and passing through the point (1, -2, 4).
Note: The equation of plane using the point (3, -4, 5) and normal is,
Hence we got the same equation of plane which is perpendicular to the plane and perpendicular to the yz-plane and through point (3, -4, 5).
Complete step-by-step answer:
It is said that a plane is perpendicular to the yz-plane, then the plane will be parallel to the x-axis. If the planes are perpendicular, then their normals are also perpendicular. Thus the normal vector, can be said as (1, 0, 0), as the normal of the x-axis, parallel to plane and yz are perpendicular.
Now we need to find the vector which is lying in the plane by using the 2 given points (1, -2, 4) and (3, -4, 5). Let us mark that vector as
Similarly, the parallel vector
So the normal of the plane is given as,
Let us take the cross product.
Therefore the equation of plane using the point (1, -2, 4) and normal
Hence we got the equation of plane perpendicular to the yz-plane and passing through the point (1, -2, 4).
Note: The equation of plane using the point (3, -4, 5) and normal
Hence we got the same equation of plane which is perpendicular to the plane and perpendicular to the yz-plane and through point (3, -4, 5).
Latest Vedantu courses for you
Grade 11 Science PCM | CBSE | SCHOOL | English
CBSE (2025-26)
School Full course for CBSE students
₹41,848 per year
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 10 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which tributary of Indus originates from Himachal Pradesh class 10 social science CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Distinguish between ordinary light and laser light class 10 physics CBSE
