Two equations $\to r ._{1} = q_{1}$ and $\to r ._{2} = q_{2}$ represent two perpendicular planes, where $_{1}$ and $_{2}$ are two unit vectors. One of these plane is rotated through an angle 45 about line of intersection of two given planes then equation of plane in new position can be

\to r . (_{1} +_{2}) = q_{1} - q_{2}

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

\to r . (_{1} +_{2}) = q_{1} + q_{2}

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

\to r . (_{1} -_{2}) = q_{1} + q_{2}

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

\to r . (_{1} -_{2}) = q_{1} - q_{2}

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

Open in App

Solution

The correct options are
B $\to r . (_{1} +_{2}) = q_{1} + q_{2}$
D $\to r . (_{1} -_{2}) = q_{1} - q_{2}$
Equation of a plane passing through intersection of two plane
$\to r ._{1} - q_{1} + λ (\to r . \to n_{2} - q_{2}) = 0$
$\to r . (_{1} + λ_{2}) = q_{1} + λ q_{2} - - - (1)$
angle between (i) & $\to r ._{1} = q$
$c o s 45^{\circ} = \frac{_{1} . (_{1} + λ_{2})}{|_{1} | |_{1} + λ_{2} |}$
$\frac{1}{\sqrt{2}} = \frac{1}{1. \sqrt{1 + λ^{2}}} \Rightarrow 1 + λ^{2} = 2$
(Here $_{1} ._{2} = 0$ and $_{1} ._{2} = 1$ )
$λ^{2} = 1, λ = \pm 1$
$\to r . (_{1} +_{2}) = q_{1} + q_{2}$
$\to r . (_{1} -_{2}) = q_{1} - q_{2}$

Suggest Corrections

Similar questions

\to r ._{1} = q_{1}

\to r ._{2} = q_{2}

P_{1} Q_{1} & P_{2} Q_{2}

y^{2} = 4 a x,

P_{1} P_{2} & Q_{1} Q_{2}

q_{1} + q_{2} = q

q_{1}

q_{2}

\frac{q_{1}}{q_{2}}

Consider the situation shown in figure. What are the signs of $q_{1} a n d q_{2}$ ?

(0, 0), (3, \sqrt{3}), (p, q)

q_{1}, q_{2}

q

q_{1} + q_{2} = ?

Join BYJU'S Learning Program