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- The angle between a- and the vector -2-2k- is

The correct option is B #10; π4 #10; P_1 is plane by î,î+î #160;r_1=î +λ ( ĵ ) P_2 in plane by î ĵ,ĵ+k̂ #160;r_2=î ĵ+μ ...

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Byju's Answer

Standard XII

Physics

Slope

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 $^i -^j,^j +^k$ . The angle between $\to a$ and the vector $^i - 2^j + 2^k$ is

\frac{π}{4}

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\frac{π}{3}

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\frac{π}{6}

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\frac{π}{2}

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Solution

The correct option is B $\frac{π}{4}$
$P_{1}$ is plane by $^i,^i +^i$
${\to r}_{1} =^i + λ (^j)$
$P_{2}$ in plane by $^i -^j,^j +^k$
${\to r}_{2} =^i -^j + μ (-^i + 2^j +^k)$
${\to r}_{1} = {\to r}_{2}$
$^i + λ^i = (1 - μ)^i + (2 μ - 1)^i + μ^k$
$μ = 0$
$λ = - 1$
So $\to a = k (^i -^j)$
$\to a - (^1 - 2^j + 2^k) = | \to a | ∣ ∣^i - 2^j + 2^k ∣ ∣ cos θ$
$\frac{k (1 + 2)}{3} = k \sqrt{2} cos θ$
$c o s θ = \frac{1}{\sqrt{2}}$
$θ = \frac{π}{4}$

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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,^j+^k. The angle between →a and the vector ^i−2^j+2^k is

The correct option is B π4 P1 is plane by ^i,^i+^i→r1=^i +λ(^j)P2 in plane by ^i−^j,^j+^k→r2=^i−^j+μ(−^i+2^j+^k)→r1=→r2^i+λ^i=(1−μ)^i+(2μ−1)^i+μ ^kμ=0λ=−1So →a=k(^i−^j)→a−(^1−2^j+2^k)=|→a|∣∣^i−2^j+2^k∣∣cosθk(1+2)3=k √2 cosθcos θ=1√2θ=π4

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 $^i -^j,^j +^k$ . The angle between $\to a$ and the vector $^i - 2^j + 2^k$ is