The cartesian from of the plane ${\bf{\vec r}} = \left( {s - 2t} \right)\widehat {{\bf{\dot i}}} + \left( {3 - t} \right)\widehat {{\bf{\dot j}}} + \left( {2s + t} \right){\bf{\hat k}}$ is

JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy

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Application +4 | -1

The cartesian from of the plane

\mathbf{\vec r} = \left( {s - 2t} \right)\widehat {\mathbf{\dot i}} + \left( {3 - t} \right)\widehat {\mathbf{\dot j}} + \left( {2s + t} \right)\mathbf{\hat k}

View answer and explanation

Correct option: C

2x - 5y - z + 15 = 0

Option C (

2x - 5y - z + 15 = 0

) is correct because it is derived by finding the normal vector to the plane and using the point-normal form of the plane equation.

Practice All Chapter MCQs →

\mathbf{r} .(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})+5=0

\mathbf{r}-(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})-8=0

\frac{1}{3}\left( {\hat i - 2\hat j + 2\hat k} \right),\frac{1}{5}\left( { - 4\hat i - 3\hat k} \right)

{\hat j}

\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

\mathbf{b}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}

\frac{\text { Projection of vector } \mathbf{a} \text { on vector } \mathbf{b}}{\text { Projection of vector } \mathbf{b} \text { on vector } \mathbf{a}}

{\mathbf{\vec a}}

{\mathbf{\vec b}}

\frac{{5\pi }}{6}

{\mathbf{\vec a}}

{\mathbf{\vec b}}

\frac{{ - 6}}{{\sqrt 3 }}

\left| {\mathbf{\vec a}} \right|

\frac{1}{4} \mathrm{~m} / \mathrm{s}

O A

O B

30^{\circ}

45^{\circ}

O B

2x + 4y - 5z = 10