A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors - - - - and the plane determined by the vectors - - - - - k- If the acute angle between a and the vector - - 2- - 2k- is - then find the value of - 2cos-

The correct option is B 1Equation of the plane containing î and î +ĵ is ⇒( r î #160;)·[ î×( î +ĵ #160;) #160;] =0 ⇒{( xî +yĵ +zk ...

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Standard XII

Mathematics

Condition for a Line to Lie on a Plane

A non-zero ve...

Question

A non-zero vector $\to a$ is parallel to the line of intersection of the plane determined by the vectors $^i,^i +^j$ and the plane determined by the vectors $^i -^j,^i +^k .$ If the acute angle between $\to a$ and the vector $^i - 2^i + 2^k$ is $θ$ , then find the value of $\sqrt{2} cos θ$ .

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Solution

The correct option is B 1
Equation of the plane containing $^i$ and $^i +^j$ is
$\Rightarrow (\to r -^i) \cdot [^i \times (^i +^j)] = 0$
$⇒{(x^i+y^j+z^k)−^i}.[^i×^i+^i×^j]=0$
$⇒{(x−1)^i+y^j+z^k}.[^k]=0$ $\Rightarrow (x - 1)^i .^k + y^j .^k + z^k .^k = 0$ $\Rightarrow z = 0$ ...(i)
Equation of the plane containing $^i -^j$ and $^i +^k$ is
$\Rightarrow (\to r -^i +^j) \cdot [(^i -^j) \times (^i +^k)] = 0$
$⇒{(x^i+y^j+z^k)−(^i−^j)}.[^i×^i+^i×^k−^j×^i−^j×^k]=0$ $⇒{(x−1)^i+(y+1)^j+z^k}.[−^j+^k−^i]=0$
$\Rightarrow - (x - 1) - (y + 1) + z = 0$
$\Rightarrow x + y - z = 0$ ...(ii)

Let $\to a = a_{1}^i + a_{2}^j + a_{3}^k$
Since $\to a$ is parallel to equations (i) and (ii), we obtain
$a_{3} = 0$ and $a_{1} + a_{2} - a_{3} = 0$
$\Rightarrow a_{1} = - a_{2}, a_{3} = 0$
Thus a vector in the direction $\to a$ is $^i -^j$
If $θ$ is the angle between $\to a$ and $^i - 2^j + 2^k$ , then
$cos θ = \pm \frac{(1) (1) + (- 1) (- 2)}{\sqrt{1 + 1} \sqrt{1 + 4 + 4}} = \pm \frac{3}{\sqrt{2} \times 3}$
$\Rightarrow \sqrt{2} cos θ = \pm 1$
$∴ \sqrt{2} cos θ = 1$ $[∵ θ is acute]$

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A non-zero vector →a is parallel to the line of intersection of the plane determined by the vectors ^i,^i+^j and the plane determined by the vectors ^i−^j,^i+^k. If the acute angle between →a and the vector ^i−2^i+2^k is θ, then find the value of √2cosθ.

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