Let a- - i- - k- b- - x i- - j- - 1-x k- and c- - y i- - x j- - 1 - x - y k- Then - a- b- c- depends on

The correct option is C Neither x nor y a .( b × c ) =| 1 amp; 0 amp; 1 x amp; 1 amp; 1 x y amp; x amp; 1+x y | ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Matrix

Let a⃗ = ...

Question

Let $\to a$ = $\to i$ - $\to k$ , $\to b$ = x $\to i$ + $\to j$ + (1-x) $\to k$ and $\to c$ = y $\to i$ + x $\to j$ + (1 + x - y) $\to k$ . Then [ $\to a$ $\to b$ $\to c$ ] depends on

only x

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only y

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Neither x nor y

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both x and y

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Solution

The correct option is C Neither x nor y
$¯ ¯ ¯ a . (¯ ¯ b \times ¯ ¯ c)$
$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & - 1 x & 1 & 1 - x y & x & 1 + x - y \end{matrix} ∣ ∣ ∣ ∣$
$⟹ 1 (1 + x - y - x + x^{2}) + 0 - 1 (x^{2} - y)$
$⟹ 1$
it depends neither on x or y.

Suggest Corrections

Similar questions

Q. If

\to a = \to i + \to j + \to k, \to a = 2 \to i + \to k, \to c = 2 \to i + \to j + \to k; \to d = \to i + \to j + 2 \to k

, then verify that

(\to a \times \to b) \times (\to c \times \to d) = [\to a \to b \to d] \to c - [\to a \to b \to c] \to d

Q. Let

\to a = \to i - \to k

\to b = x \to i + \to j + (1 - x) \to k

and

\to c = y \to i + x \to j + (1 + x - y) \to k

, then

\to a, \to b

and

\to c

are non-coplanar for

Q. Let

\to a = \to i - \to k

\to b = x \to i + \to j + (1 - x) \to k

and

\to c = y \to i + x \to j + (1 + x - y) \to k

, then

[\to a \to b \to c]

depends on

Q. Let

\to a = \to i + \to j + \to k, \to c = \to j - \to k

. If

\to b

is a vector satisfying

\to a \times \to b = \to c

and

\to a . \to b = 3

, then

\to b

Q. Three vectors

\to A = a \to i + \to j + \to k, \to B = \to i + b \to j + \to k, \to C = \to i + \to j + c \to k

are mutually perpendicular (

\to i, \to j, \to k

are unit vectors along

X, Y, Z

axis respectively). The respective values of

a, b

and

c

are

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Let →a = →i - →k, →b = x→i + →j + (1-x)→k and →c = y→i + x→j + (1 + x - y)→k. Then [→a →b →c] depends on

The correct option is C Neither x nor y¯¯¯a.(¯¯b×¯¯c)=∣∣ ∣∣10−1x11−xyx1+x−y∣∣ ∣∣⟹1(1+x−y−x+x2)+0−1(x2−y)⟹1it depends neither on x or y.

Let $\to a$ = $\to i$ - $\to k$ , $\to b$ = x $\to i$ + $\to j$ + (1-x) $\to k$ and $\to c$ = y $\to i$ + x $\to j$ + (1 + x - y) $\to k$ . Then [ $\to a$ $\to b$ $\to c$ ] depends on

The correct option is C Neither x nor y
$¯ ¯ ¯ a . (¯ ¯ b \times ¯ ¯ c)$
$= ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & - 1 x & 1 & 1 - x y & x & 1 + x - y \end{matrix} ∣ ∣ ∣ ∣$
$⟹ 1 (1 + x - y - x + x^{2}) + 0 - 1 (x^{2} - y)$
$⟹ 1$
it depends neither on x or y.