The vector equation of a plane which con...

The vector equation of a plane which contains the line `vecr=2hati+lamda(hatj-hatk)` and perpendicular to the plane `vecr.(hati+hatk)=3` is

`vecr.(hati-hatj-hatk)=2`

`vecr.(hati+hatj-hatk)=2`

`vecr.(hati+hatj+hatk)=2`

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Let A be the given point whose position vector relative to an origin O be veca and vec(ON)=vecn. let vecr be the position vector of any point P which lies on the plane passing through A and perpendicular to ON.Then for any point P on the plane, vec(AP).vecn=0 or, (vecr-veca).vecn=0 or vecr.vecn=veca.vecn or , vecr.hatn=p where p ils the perpendicular distance of the plane from origin. The equation fo the plane which ponts contains the line vecr=2hati+t(hatj-hatk0 where it is scalar and perpendicular to the plane vecr.(hati-hatk)=3 is (A) vecr.(hati-hatj-hatk)=2 (B) vecr.(hati+hatj-hatk)=2 (C) vecr.(hati+hatj+hatk)=2 (D) vecr.(2hati+hatj-hatk)=4

The equation of the plane containing the line vecr=hati+hatj+lamda(2hati+hatj+4hatk) is

Find the vector equation to the plane through the point -hati+3hatj+2hatk perpendicular to each of the planes vecr.(hati+2hatj+2hatk)=25 and vecr.(3hati+3hatj+2hatk)=8.

Angle between the line vecr=(2hati-hatj+hatk)+lamda(-hati+hatj+hatk) and the plane vecr.(3hati+2hatj-hatk)=4 is

The vector equation of the plane containing he line vecr=(-2hati-3hatj+4hatk)+lamda(3hati-2hatj-hatk) and the point hati+2hatj+3hatk is

Find the vector equation of the plane in which the lines vecr=hati+hatj+lambda(hati+2hatj-hatk) and vecr=(hati+hatj)+mu(-hati+hatj-2hatk) lie.

Find the unit vector perpendicular to the plane vecr*(2hati+hatj+2hatk)=5 .

The equation of the plane containing the lines vecr=(hati+hatj-hatk)+lamda(3hati-hatj) and vecr=(4hati-hatk)+mu(2hati+3hatk) , is

Find the equation of the plane through the line of intersection of the planes vecr.(2hati-3hatj+4hatk)=1 and vecr.(hati-hatj)+4=0 and perpendicular to the plane vecr.(2hati-hatj+hatk)+8=0 .

OBJECTIVE RD SHARMA ENGLISH-PLANE AND STRAIGHT LINE IN SPACE -Exercise

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