If a-2i-k-b-i-j-k-c-4i-3j-7k- then the vector r satisying r-b- c-b and r-a-0 is

The correct option is D i 8j+2k i 8j+2k r× nbsp;b c×b=0 ( r c )×b=0 ( r c )// b r c=tb r=c+tb r a=c.a+t(b.a) ...

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Byju's Answer

Standard XII

Mathematics

Multiplication of a Vector by a Scalar

If a=2i+k,b...

Question

If $¯ ¯ ¯ a = 2 ¯ i + ¯ ¯ ¯ k, ¯ ¯ b = ¯ i + ¯ j + ¯ ¯ ¯ k, ¯ ¯ c = 4 ¯ i - 3 ¯ j + 7 ¯ ¯ ¯ k,$ then the vector $¯ ¯ ¯ r$ satisying $¯ ¯ ¯ r \times ¯ ¯ b = ¯ ¯ c \times ¯ ¯ b$ and $¯ ¯ ¯ r . ¯ ¯ ¯ a = 0$ is

¯ i + 8 ¯ j + 2 ¯ ¯ ¯ k

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¯ i - 8 ¯ j + 2 ¯ ¯ ¯ k

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¯ i - 8 ¯ j - 2 ¯ ¯ ¯ k

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- ¯ i - 8 ¯ j + 2 ¯ ¯ ¯ k

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Solution

The correct option is D $- ¯ i - 8 ¯ j + 2 ¯ ¯ ¯ k$
$- ¯ i - 8 ¯ j + 2 ¯ ¯ ¯ k$
$¯ ¯ ¯ r \times ¯ ¯ b - ¯ ¯ c \times ¯ ¯ b = ¯ ¯ ¯ 0$
$(¯ ¯ ¯ r - ¯ ¯ c) \times ¯ ¯ b = ¯ ¯ ¯ 0$
$(¯ ¯ ¯ r - ¯ ¯ c) / / ¯ ¯ b$
$¯ ¯ ¯ r - ¯ ¯ c = t ¯ ¯ b$
$¯ ¯ ¯ r = ¯ ¯ c + t ¯ ¯ b$
$¯ ¯ ¯ r - ¯ ¯ ¯ a = ¯ ¯ c . ¯ ¯ ¯ a + t (¯ ¯ b . ¯ ¯ ¯ a)$
$\frac{- (¯ ¯ c . ¯ ¯ ¯ a)}{¯ ¯ b . ¯ ¯ ¯ a} = t$
$\to r = \to c - \frac{\to c . \to a}{\to b . \to a}$

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Similar questions

Q. A plane passing through the point(-1,1,1) is parallel to the vector

2 ˆ i + 3 ˆ j - 7 ˆ k

and line

r = (ˆ i - 2 ˆ j - ˆ k) + λ (3 ˆ i - 8 ˆ j + 2 ˆ k)

. Find the vector equation of the plane.

Q. Find

\vec{a} \cdot \vec{b}

when
(i)

\vec{a} = \hat{i} - 2 \hat{j} + \hat{k} and \vec{b} = 4 \hat{i} - 4 \hat{j} + 7 \hat{k}

(ii)

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

(iii)

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

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 lines whose vector equations are:

(i)

\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + λ (3 \hat{i} - \hat{j} + \hat{k}) and \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})

(ii)

\vec{r} = (3 \hat{i} + 5 \hat{j} + 7 \hat{k}) + λ (\hat{i} - 2 \hat{j} + 7 \hat{k}) and \vec{r} = - \hat{i} - \hat{j} - \hat{k} + μ (7 \hat{i} - 6 \hat{j} + \hat{k})

(iii)

\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) and \vec{r} = (2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + μ (3 \hat{i} + 4 \hat{j} + 5 \hat{k})

(iv)

\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - t) \hat{k} and \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}

(v)

\vec{r} = (λ - 1) \hat{i} + (λ + 1) \hat{j} - (1 + λ) \hat{k} and \vec{r} = (1 - μ) \hat{i} + (2 μ - 1) \hat{j} + (μ + 2) \hat{k}

(vi)

\vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + λ (2 \hat{i} - 5 \hat{j} + 2 \hat{k}) and, \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})

(vii)

\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 (3 \hat{i} - 5 \hat{j} + 2 \hat{k})

(viii)

\vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k}

and

\vec{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})

[NCERT EXEMPLAR]

Q. If

A = 2 i + k, B = i + j + k

and

C = 4 i - 3 j + 7 k

, then a vector

r

which satisfies

R \times B = C \times B

and

R . A = 0

, is

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If ¯¯¯a=2¯i+¯¯¯k,¯¯b=¯i+¯j+¯¯¯k,¯¯c=4¯i−3¯j+7¯¯¯k, then the vector ¯¯¯r satisying ¯¯¯r×¯¯b=¯¯c×¯¯b and ¯¯¯r.¯¯¯a=0 is

If $¯ ¯ ¯ a = 2 ¯ i + ¯ ¯ ¯ k, ¯ ¯ b = ¯ i + ¯ j + ¯ ¯ ¯ k, ¯ ¯ c = 4 ¯ i - 3 ¯ j + 7 ¯ ¯ ¯ k,$ then the vector $¯ ¯ ¯ r$ satisying $¯ ¯ ¯ r \times ¯ ¯ b = ¯ ¯ c \times ¯ ¯ b$ and $¯ ¯ ¯ r . ¯ ¯ ¯ a = 0$ is