Find a vector r- in the plane of p- and q-k- such that r- is perpendicular to p- and r-q-2

We have, r #160; = xp #160; + yq = x( i #160; + j) + y( j #160; + k) r = x i +( x y ) j +y k #160; r . p =0 r ...

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Byju's Answer

Standard XII

Mathematics

Family of Planes Passing through the Intersection of Two Planes

Find a vector...

Question

Find a vector $\to r$ in the plane of $\to p = -^i +^j$ and $\to q = -^j +^k$ such that $\to r$ is perpendicular to $\to p$ and $\to r . \to q = - 2$

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Solution

We have,
$¯ ¯ ¯ r = x ¯ ¯ ¯ p + y ¯ ¯ ¯ q$
$= x (- ¯ i + ¯ j) + y (- ¯ j + ¯ ¯ ¯ k)$
$\begin{matrix} ¯ ¯ ¯ r = - x ¯ i + (x - y) ¯ j + y ¯ ¯ ¯ k ¯ ¯ ¯ r . ¯ ¯ ¯ p = 0 ¯ ¯ ¯ r . ¯ ¯ ¯ p = - 2 [- x ¯ i + (x - y) ¯ j + y ¯ ¯ ¯ k] . (- ¯ i + ¯ j) = 0 x + x - y = 0 2 x = y ¯ ¯ ¯ r . ¯ ¯ ¯ q = - 2 y - x + y = - 2 2 y - x = - 2 4 x - x = - 2 x = - \frac{2}{3} y = - \frac{4}{3} ¯ ¯ ¯ r = - \frac{2}{3} ¯ ¯ ¯ p - \frac{4}{3} ¯ ¯ ¯ q \end{matrix}$

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Similar questions

Q. Find a vector

\to r

in the plane of

\to p = -^i +^j

and

\to q = -^j +^k

such that

\to r

is perpendicular to

\to p

and

\to r . \to q = - 2

Q. Three vectors

\to P

\to Q

\to R

are such that the

| \to P |

= | \to Q |

| \to R |

= \sqrt{2}

| \to P |

and

\to P + \to Q

+ \to R = 0

. The angle between

\to P

and

\to Q

\to Q

and

\to R

and

\to P

and

\to R

will be respectively.

Q. If

[\to p + 2 \to q + 3 \to r \to q + 2 \to r + 3 \to p \to r + 2 \to p + 3 \to q] = 54

where

\to p

\to q

and

\to r

are three vector then the Value of

∣ ∣ ∣ ∣ \begin{matrix} \to p . \to p & \to p . \to q & \to p . \to r \to p . \to q & \to q . \to q & \to q . \to r \to p . \to r & \to r . \to q & \to r . \to r \end{matrix} ∣ ∣ ∣ ∣

Q. If

\to q, \to r

are unit vectors such that

\to p = \to q \times \to p + \to r

, then

Q. lf

\to P \times \to Q = \to R; \to Q \times \to R = \to P

and

\to R \times \to P = \to Q

are 3 non-zero vectors, then,
a)

\to P, \to Q

and

\to R

are coplanar
b) Angle between

\to P

and

\to Q

may be less than

90^{0}

\to P + \to Q + \to R

cannot be equal to zero
d)

\to P, \to Q

and

\to R

are mutually perpendicular
( Given

P \neq Q \neq R \neq 0

)

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Find a vector →r in the plane of →p=−^i+^j and →q=−^j+^k such that →r is perpendicular to →p and →r.→q=−2

We have,¯¯¯r=x¯¯¯p+y¯¯¯q=x(−¯i+¯j)+y(−¯j+¯¯¯k)¯¯¯r=−x¯i+(x−y)¯j+y¯¯¯k¯¯¯r.¯¯¯p=0¯¯¯r.¯¯¯p=−2[−x¯i+(x−y)¯j+y¯¯¯k].(−¯i+¯j)=0x+x−y=02x=y¯¯¯r.¯¯¯q=−2y−x+y=−22y−x=−24x−x=−2x=−23y=−43¯¯¯r=−23¯¯¯p−43¯¯¯q

Find a vector $\to r$ in the plane of $\to p = -^i +^j$ and $\to q = -^j +^k$ such that $\to r$ is perpendicular to $\to p$ and $\to r . \to q = - 2$