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Byju's Answer

Standard XII

Mathematics

Angle between Two Planes

Find the angl...

Question

Find the angles between the following pairs of vectors: $i - j + k, - i + j + 2 k,$
and $i - j + k, - i + j + 2 k$

\frac{π}{6}

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\frac{π}{2}

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\frac{π}{3}

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\frac{π}{4}

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Solution

The correct option is B $\frac{π}{2}$
Dot product of two vectors $\to a = x_{1} i + y_{1} j + z_{1} k$ and $\to b = x_{2} i + y_{2} j + z_{2} k$ = $\to a . \to b$ $= x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}$ $= | \to a | ∣ ∣ \to b ∣ ∣ c o s θ$
Where $θ$ is angle between two vectors
Given $i - j + k$ , $- i + j + 2 k$
$(i - j + k) . (- i + j + 2 k) = 0 = \sqrt{3} . \sqrt{6} c o s θ$
$\Rightarrow c o s θ = 0 \Rightarrow θ = \frac{π}{2}$

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Similar questions

Q. Find the angle between the vectors

\vec{a} and \vec{b}

, where
(i)

\vec{a} = \hat{i} - \hat{j} and \vec{b} = \hat{j} + \hat{k}

(ii)

\vec{a} = 3 \hat{i} - 2 \hat{j} - 6 \hat{k} and \vec{b} = 4 \hat{i} - \hat{j} + 8 \hat{k}

(iii)

\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} and \vec{b} = 4 \hat{i} + 4 \hat{j} - 2 \hat{k}

(iv)

\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} and \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}

(v)

\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}

Q. Find the shortest distance between the following pairs of parallel lines whose equations are:
(i)

\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) and \vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + μ (- \hat{i} + \hat{j} - \hat{k})

(ii)

\vec{r} = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k}) and \vec{r} = (2 \hat{i} + \hat{j} - \hat{k}) + μ (4 \hat{i} - 2 \hat{j} + 2 \hat{k})

Q. The vectors

λ i + j + 2 k, i + λ j - k

and

2 i - j + λ k

are coplanar if

Let $I = \int_{0}^{\frac{π}{6}} \frac{c o s x}{x} d x, J = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{c o s x}{x} d x .$ Which of the following is correct?

Q. A unit vectors coplanar with

i + j + 2 k

and

i + 2 j + k

and perpendicular to

i + j + k

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Find the angles between the following pairs of vectors: i−j+k,−i+j+2k,and i−j+k,−i+j+2k

The correct option is B π2Dot product of two vectors →a=x1i+y1j+z1k and →b=x2i+y2j+z2k = →a.→b=x1x2+y1y2+z1z2=|→a|∣∣→b∣∣cosθWhere θ is angle between two vectorsGiven i−j+k, −i+j+2k(i−j+k).(−i+j+2k)=0=√3.√6cosθ⇒cosθ=0⇒θ=π2

Find the angles between the following pairs of vectors: $i - j + k, - i + j + 2 k,$
and $i - j + k, - i + j + 2 k$