You visited us 99 times! Enjoying our articles? Unlock Full Access!

Coordinate Axes in Three Dimensional Space

The non-param...

Question

The non-parametric vector equation of a plane passing through a point whose position vector is $\to a$ is parallel to $\to u$ and $\to v$ is

[\to r - \to a, \to u, \to v] = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[\to r, \to u, \to v] = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[\to r, \to a, \to u \times \to v]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[\to a, \to u, \to v] = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $[\to r - \to a, \to u, \to v] = 0$
Lets recall non-parametric vector equation.
Suppose we want to define a plane that passes through $P_{0}$ and is perpendicular to n.
NOTE: There will be only one such plane because fixing a direction fixes the orientation of our plane.

Lets define r as set of all points which lie on the plane.
Now, observe that $r - r_{0}$ lies on the plane itself because both
$P, P_{0}$ lie on the plane.
Thus, equation of plane becomes
$(r - r_{0}) . n = 0$ (because $c o s 90 = 0$ )

In the given case , $n = u \times v$ and $r_{0} = a$
Replacing in our equation we get, $(r - a) . (u \times v) = 0$
$[r - a, u, v] = 0$

Suggest Corrections

Similar questions

\to u = \to a - \to b, \to v = \to a + \to b

| \to a | = | \to b | = 2

| \to u \times \to v |

$I f \to u = \to a - \to b a n d \to v = \to a + \to b a n d | \to a | = | \to b | = 2, t h e n | \to u \times \to v | =$

\to u = \to a - \to b; \to v = \to a + \to b

| \to a | = ∣ ∣ \to b ∣ ∣ = 2

| \to u \times \to v |

\to a

\to b

\to u = \to a - (\to a . \to b) \to b

\to v = \to a \times \to b,

| \to v |

\to u

\to v

\to w

| \to u |

| \to v |

| \to w |

\to v

\to u

\to w

\to u

\to v

\to w

| \to u \to v + \to w |

Join BYJU'S Learning Program